Rd Sharma 2021 for Class 9 Maths Chapter 11 - Triangle And Its Angles

Answer:

$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°.\right] \\ \Rightarrow 55°+40°+\angle C=180° \\ \Rightarrow 95°+\angle C=180° \\ \Rightarrow \angle C=180°-95° \\ \Rightarrow \angle C=85° \end{gathered}$$
 

Answer:

Let the angles of the given triangle be of xº, 2xº and 3xº. Then,

$$\begin{gathered} \therefore x+2x+3x=180 \left[\mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 6x=180 \\ \Rightarrow x=30 \end{gathered}$$

Answer:

Given angles are

$$\begin{gathered} \therefore \left(x-40\right)+\left(x-20\right)+\left(\frac{1}{2}x-10\right)=180 \\ \Rightarrow \frac{5}{2}x=180+70 \\ \Rightarrow \frac{5}{2}x=250 \\ \Rightarrow x=\frac{250\times 2}{5} \\ \Rightarrow x=100 \end{gathered}$$

Hence, the value of x is 100°.

Answer:

Let the two equal angles are x°, then the third angle will be (x + 30)°.

Therefore, the angles of the given triangle are 50°, 50° and 80°.

Answer:

Let ABC be a triangle such that

$$\begin{gathered} \angle A=\angle B+\angle C \left[\mathrm{Since}, \mathrm{one} \mathrm{angle} \mathrm{is} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{two}\right] \\ \therefore \angle A+\angle B+\angle C=180° \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow \angle A+\angle A=180° \\ \Rightarrow 2\angle A=180° \\ \Rightarrow \angle A=90° \end{gathered}$$
 

Answer:

(i) Let a triangle ABC has two angles equal to .  We know that sum of the three angles of a triangle is 180°.

Hence, if two angles are equal to , then the third one will be equal to zero which implies that A, B, C is collinear, or we can say ABC is not a triangle

A triangle can’t have two right angles.

(ii) Let a triangle ABC has  two obtuse angles

This implies that sum of only two angles will be equal to more than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a  triangle can’t have two obtuse angles.

(iii) Let a triangle ABC has two acute angles.

This implies that sum of two angles will be less than . Hence third angle will be the difference of 180° and sum of both acute angles

Therefore,  a triangle can have two acute angles.

(iv) Let a triangle ABC having angles are more than 60°.

This implies that the sum of three angles will be more than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can’t have all angles more than .

(v) Let a triangle ABC having anglesare less than 60°.

This implies that the sum of three angles will be less than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore,  a triangle can’t have all angles less than 60°.

(vi) Let a triangle ABC having angles all equal to 60°.

This implies that the sum of three angles will be equal to 180° which satisfies the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can have all angles equal to 60°.

Answer:

Let the angles of a triangle are          [Since, the difference between two consceutive angles is 10°]         

$$\begin{gathered} \therefore x+\left(x+10\right)+\left(x+20\right)=180 \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 3x+30=180 \\ \Rightarrow 3x=150 \\ \Rightarrow x=50 \end{gathered}$$

Therefore, the angles of the given triangle are 50°, (50 + 10)° and (50 + 20)° i.e. 50°, 60° and 70°.

Answer:

Since OB and OC are the angle bisector of $\angle B \mathrm{and} \angle C$

$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow 72°+2\angle OBC+2\angle OCB=180° \left[\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 2\left(\angle OBC+\angle OCB\right)=108° \\ \Rightarrow \angle OBC+\angle OCB=54° \\ \Rightarrow 180°-\angle BOC=54° \left[\mathrm{Since}, \angle OBC+\angle OCB+\angle BOC=180°\right] \\ \Rightarrow \angle BOC=126° \end{gathered}$$

Hence magnitude of $\angle BOC \mathrm{is} 126°.$

Answer:

Let ABC be a triangle and  BO and CO be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠ABC and ∠ACB of a triangle ABC meet at a point O, then

$\angle BOC=90°+\frac{1}{2}\angle A$

From the above relation it is very clear that if is equals 90° then must be equal to zero.

Now, if possible let is equals zero but on other hand it represents that  A, B, C will be collinear, that is they do not form a triangle.

It leads to a contradiction.

Hence, the bisectors of base angles of a triangle cannot enclose a right angle in any case.

Answer:

Let ABC be a triangle and Let BO and CO be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠ABC and ∠ACB of a triangle ABC meet at a point O, then

$\angle BOC=90°+\frac{1}{2}\angle A$

Hence the triangle is a right angled triangle.

Answer:

Let ABC be a triangle and BO and CO be the bisectors of the base anglerespectively.

$$\begin{gathered} \therefore 120°=90°+\frac{1}{2}\angle A \\ \Rightarrow 30°=\frac{1}{2}\angle A \\ \Rightarrow \angle A=60° \end{gathered}$$

are equal as it is given that .

$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\mathrm{Sum} \mathrm{of} \mathrm{three} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \Rightarrow 60°+2\angle B=180° \left[\because \angle ABC=\angle ACB\right] \\ \Rightarrow \angle B=60° \end{gathered}$$

Hence, .

Answer:

Let a triangle ABC having angles.

It is given that the sum of two angles are less than third one.

We know that the sum of all angles of a triangle equal to 180°.

Similarly we can prove that

Since,  all angles are less than 90°.
Hence,  triangle is acute angled.

Answer:

In the given problem, the exterior angles obtained on producing the base of a triangle both ways are and . So, let us draw ΔABC and extend the base BC, such that:

Here, we need to find all the three angles of the triangle.

Now, since BCD is a straight line, using the property, “angles forming a linear pair are supplementary”, we get

Similarly, EBC is a straight line, so we get,

Further, using angle sum property in ΔABC

Therefore,.

Answer:

In the given ΔABC, and . We need to find .

Here, are vertically opposite angles. So, using the property, “vertically opposite angles are equal”, we get,

Further, BCD is a straight line. So, using linear pair property, we get,

Now, in ΔABC, using “the angle sum property”, we get,

Therefore,.

Answer:

In the given problem, we need to find the value of x

(i) In the given ΔABC, and

Now, BCD is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Similarly, EAC is a straight line. So, we get,

Further, using the angle sum property of a triangle,

In ΔABC

Therefore,

(ii) In the given ΔABC, and

Here, BCD is a straight line. So, using the property, “the angles forming a linear pair are supplementary” we get,

Similarly, EBC is a straight line. So, we get

Further, using the angle sum property of a triangle,

In ΔABC

Therefore,

(iii) In the given figure,and

Here,and AD is the transversal, so form a pair of alternate interior angles. Therefore, using the property, “alternate interior angles are equal”, we get,

Further, applying angle sum property of the triangle

In ΔDEC

Therefore,

Answer:

In the given figure,and. We need to find the value of

Since, 

Let, 

Applying the angle sum property of the triangle, in ΔABC, we get,

Thus,

Further, BCD is a straight line. So, applying the property, “the angles forming a linear pair are supplementary”, we get,

Therefore,.

Answer:

In the given problem,

We need to find

Now,and AE is the transversal, so using the property, “alternate interior angles are equal”, we get,

Further, applying angle sum property of the triangle

In ΔDCE

Further, ACE is a straight line, so using the property, “the angles forming a linear pair are supplementary”, we get,

Therefore,.

Answer:

(i) Sum of the three angles of a triangle is 180°

According to the angle sum property of the triangle

In ΔABC

Hence, the given statement is.

(ii) A triangle can have two right angles.

According to the angle sum property of the triangle

In ΔABC

Now, if there are two right angles in a triangle

Let

Then, 

(This is not possible.)

Therefore, the given statement is.

(iii) All the angles of a triangle can be less than 60°

According to the angle sum property of the triangle

In ΔABC

Now, If all the three angles of a triangle is less than

Then, 

Therefore, the given statement is.

(iv) All the angles of a triangle can be greater than 60°

According to the angle sum property of the triangle

In ΔABC

Now, if all the three angles of a triangle is greater than

Then, 

Therefore, the given statement is.

(v) All the angles of a triangle can be equal to

According to the angle sum property of the triangle

In ΔABC

Now, if all the three angles of a triangle are equal to

Then, 

Therefore, the given statement is.

(vi) A triangle can have two obtuse angles.

According to the angle sum property of the triangle

In ΔABC

Now, if a triangle has two obtuse angles

Then, 

Therefore, the given statement is.

(vii) A triangle can have at most one obtuse angle.

According to the angle sum property of the triangle

In ΔABC

Now, if a triangle will have more than one obtuse angle

Then, 

Therefore, the given statement is.

(viii) If one angle of a triangle is obtuse, then it cannot be a right angles triangle.

According to the angle sum property of the triangle

In ΔABC

Now, if it is a right angled triangle

Then, 

Also if one of the angle’s is obtuse

This is not possible.

Thus, if one angle of a triangle is obtuse, then it cannot be a right angled triangle.

Therefore, the given statement is.

(ix) An exterior angle of a triangle is less than either of its interior opposite angles

According to the exterior angle property, an exterior angle of a triangle is equal to the sum of the two opposite interior angles.

In ΔABC

Let x be the exterior angle

So,

Now, if x is less than either of its interior opposite angles

Therefore, the given statement is.

(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

According to exterior angle theorem,

Therefore, the given statement is.

(xi) An exterior angle of a triangle is greater than the opposite interior angles.

According to exterior angle theorem,

Since, the exterior angle is the sum of its interior angles.

Thus, 

Therefore, the given statement is.

Answer:

(i) Sum of the angles of a triangle is 180°.

As we know, that according to the angle sum property, sum of all the angles of a triangle is 180°.

(ii) An exterior angle of a triangle is equal to the two interior opposite angles.

(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.

As according to the property: An exterior angle of a triangle is equal to the sum of two interior opposite angles. Therefore, it has to be greater than either of them.

(iv) A triangle cannot have more than one right angle.

As the sum of all the angles of a triangle is 180°. So, if the triangle has more than one right angle the sum would exceed 180 °.

(v) A triangle cannot have more than one obtuse angle

As the sum of all the angles of a triangle is 180°. So, if the triangle has more than one obtuse angle the sum would exceed 180 °.

Answer:

In the given problem, BP and CP are the internal bisectors of respectively. Also, BQ and CQ are the external bisectors of respectively. Here, we need to prove:

We know that if the bisectors of anglesand of ΔABC meet at a point O then .

Thus, in ΔABC

              ……(1)

Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and the external bisectors of and meet at O, then”.

Thus, ΔABC

$\angle BQC=90°-\frac{1}{2}\angle A ......\left(2\right)$                

Adding (1) and (2), we get

Thus,

Hence proved.

Answer:

In the given figure,, and

Here, we will produce AD to meet BC at E

Now, using angle sum property of the triangle

In ΔAEB

Further, BEC is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Also, using the property, “an exterior angle of a triangle is equal to the sum of its two opposite interior angles”

In ΔDEC, x is its exterior angle

Thus,



Therefore,.

Answer:

In the given figure,and

Since,and angles opposite to equal sides are equal. We get,

$\angle BDA=\angle BAD .....\left(1\right)$

Also, EAD is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Further, it is given  AB divides in the ratio 1 : 3.

So, let

$\angle DAB=y, \angle BAC=3y$

Thus, 

$$\begin{gathered} y+3y=\angle DAC \\ \Rightarrow 4y=72° \\ \Rightarrow y=\frac{72°}{4} \\ \Rightarrow y=18° \end{gathered}$$

Hence, $\angle DAB=18°, \angle BAC=3\times 18°=54°$

Using (1)

Now, in ΔABC , using the property, “exterior angle of a triangle is equal to the sum of its two opposite interior angles”, we get,

$$\begin{gathered} \angle EAC=\angle ADC+x \\ \Rightarrow 108°=18°+x \\ \Rightarrow x=90° \end{gathered}$$

Therefore,.

Answer:

In the given ΔABC, the bisectors of and intersect at D

We need to prove:

Now, using the exterior angle theorem,

$\angle ABE=\angle BAC+\angle ACB$   .….(1)

 $\mathrm{As} \angle ABE \mathrm{and} \angle ACB \mathrm{are} \mathrm{bisected}$

$\angle DCB=\frac{1}{2}\angle ACB$

Also,

$\angle DBA=\frac{1}{2}\angle ABE$

Further, applying angle sum property of the triangle

In  ΔDCB

$$\begin{gathered} \angle CDB+\angle DCB+\angle CBD=180° \\ \Rightarrow \angle CDB+\frac{1}{2}\angle ACB+\left(\angle DBA+\angle ABC\right)=180° \end{gathered}$$

  $\angle CDB+\frac{1}{2}\angle ACB+\left(\frac{1}{2}\angle ABE+\angle ABC\right)=180° .....\left(2\right)$

Also, CBE is a straight line, So, using linear pair property

$$\begin{gathered} \Rightarrow \angle ABC+\angle ABE=180° \\ \Rightarrow \angle ABC+\frac{1}{2}\angle ABE+\frac{1}{2}\angle ABE=180° \\ \Rightarrow \angle ABC+\frac{1}{2}\angle ABE=180°-\frac{1}{2}\angle ABE .....\left(3\right) \end{gathered}$$

So, using (3) in (2)

$$\begin{gathered} \angle CDB+\frac{1}{2}\angle ACB+\left(180°-\frac{1}{2}\angle ABE\right)=180° \\ \Rightarrow \angle CDB+\frac{1}{2}\angle ACB-\frac{1}{2}\angle ABE=0 \\ \Rightarrow \angle CDB=\frac{1}{2}\left(\angle ABE-\angle ACB\right) \\ \Rightarrow \angle CDB=\frac{1}{2}\angle CAB \\ \Rightarrow \angle D=\frac{1}{2}\angle A \end{gathered}$$

Hence proved.

Answer:

In the given ΔABC,, is the bisector of , and

We need to find

Now, using the angle sum property of the triangle

In ΔAMC, we get,

…….(1)

Similarly,

In ΔABM, we get,

…..(2)

So, adding (1) and (2)

Now, since AN is the bisector of

Thus, 

Now,

Therefore,.

Answer:

In the given ΔABC, AD bisects and. We need to prove.

Let,

Also, 

As AD bisects, 

…..(1) 

Now, in ΔABD, using exterior angle theorem, we get,

Similarly,

[using (1)]

Further, it is given,

Adding to both the sides

Thus,

Hence proved.

Answer:

In the given ΔABC,and .

We need prove

Here, in ΔBDC, using the exterior angle theorem, we get,

Similarly, in ΔEBC, we get,

Adding (1) and (2), we get,

Now, on using angle sum property, 

In ΔABC, we get,

This can be written as,

Similarly, using angle sum property in ΔOBC, we get,

This can be written as,

Now, using the values of (4) and (5) in (3), we get,

Therefore,.

Hence proved

Answer:

In the given problem, AE bisectsand

We need to prove

As,is bisected by AE

=2=2          ..........(1)

Now, using the property, “an exterior angle of a triangle in equal to the sum of the two opposite interior angles”, we get,

()

(using 1)

Hence, using the property, if alternate interior angles are equal, then the two lines are parallel, we get,

Thus,

Hence proved.

Answer:

In a given ΔABC we are given that the three angles are equal. So,

According to the angle sum property of a triangle, in ΔABC

Therefore, all the three angles of the triangle are equal to

So, the correct option is (c).

Answer:

In the given problem, we have a right angled triangle and the other two angles are equal.

So, In ΔABC

Now, using the angle sum property of the triangle, in ΔABC, we get,

()

Therefore, the correct option is (b).

Answer:

In the ΔABC, CD is the ray extended from the vertex C of ΔABC. It is given that the exterior angle of the triangle is and two of the interior opposite angles are equal.

So, and_.

So, now using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get.

In ΔABC

Therefore, each of the two opposite interior angles is

So, the correct option is (d).

Answer:

In the given problem, one angle of a triangle is equal to the sum of the other two angles.

Thus, 

     ..........(1)

Now, according to the angle sum property of the triangle

In ΔABC

                  .........(2)

Further, using (2) in (1),

Thus,

Therefore, the correct option is (d).

Answer:

In the given problem, side BC of ΔABC has been produced to a point D. Such that and. Here, we need to find

Given

We get,

Now, using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,

In ΔABC

Also, (Using 1)

Thus,

Therefore, the correct option is (a).

Answer:

In the given ΔABC, . D is the ray extended from point A. AX bisectsand

Here, we need to find

As ray AX bisects

Thus,

Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,

Thus,

Therefore, the correct option is (c).

Answer:

In the given ΔABC, and

Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,

So,

Therefore, the correct option is (c).

Answer:

In the given ΔABC, all the three sides of the triangle are produced. We need to find the sum of the three exterior angles so produced.

Now, according to the angle sum property of the triangle

        .......(1)

Further, using the property, “an exterior angle of the triangle is equal to the sum of two opposite interior angles”, we get,

            ......(2)

Similarly,

           .......(3)

Also, 

           .......(4)

Adding (2) (3) and (4)

We get,

Thus, 

Therefore, the correct option is (d).

Answer:

In the given ΔABC,, AD bisects and .

Here, we need to find.

As, AD bisects,

We get, 

Now, according to angle sum property of the triangle

In ΔABD

Hence,

Therefore, the correct option is (c).

Answer:

In the given ΔABC, an exterior angle and its interior opposite angles are in the ratio 4:5.

Let us take,

Now using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”

We get,

Thus, 

Also, using angle sum property in ΔABC

Thus, 

Therefore, the correct option is (a).

Answer:

In the given ΔABC,and . Bisectors of and meet at O. 

We need to find

Since, OB is the bisector of.

Thus,

Now, using the angle sum property of the triangle

In ΔABC, we get,

Similarly, in ΔBOC

Hence,

Therefore, the correct option is (b).

Answer:

In the given problem, line segment AB and CD intersect at O, such that,and .

We need to find

As

Applying the property, “alternate interior angles are equal”, we get,

    .......(1)

Now, using the angle sum property of the triangle

In ΔODB, we get,

(using 1)

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure,,and . We need to find.

Here, and CD is the transversal, so using the property, “corresponding angles are equal”, we get

Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, in ΔOBD, we get, 

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, we need to find

Here, AB and CD are straight lines intersecting at point O, so using the property, “vertically opposite angles are equal”, we get,

Further, applying the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, in ΔAOC, we get,

Similarly, in ΔBOD

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, measures of the angles of ΔABC are in the ratio. We need to find the measure of the smallest angle of the triangle.

Let us take,

Now, applying angle sum property of the triangle in ΔABC, we get,

Substituting the value of x in,and

Since, the measure of is the smallest

Thus, the measure of the smallest angle of the triangle is

Therefore, the correct option is (c).

Answer:

In the given figure,

We need to find the value of x.

Now, since AB and CD are straight lines intersecting at point O, using the property, “vertically opposite angles are equal”, we get,

Further, applying angle sum property of the triangle 

In ΔBOC 

Then, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get,

In ΔEOC

Further solving for x, we get,

Thus,

Therefore, the correct option is (b).

Answer:

In the given ΔABC, we need to convert z in terms of x and y 

Now, BC is a straight line, so using the property, “angles forming a linear pair are supplementary”

Similarly,

Also, using the property, “vertically opposite angles are equal”, we get,

Further, using angle sum property of the triangle

Thus,

Therefore, the correct option is (b).

Answer:

In the given problem, we need to find the value of x if

Here, if , then using the property, “if the two lines are parallel, then the alternate interior angles are equal”, we get,

Further, applying angle sum property of the triangle

In ΔABC

Thus,

Therefore, the correct option is (d).

Answer:

In the given figure, we need to find y in terms of x

Now, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get

In ΔABC

                          ..........(1)

Similarly, in ΔOCD

(using 1)

Thus,

Therefore, the correct option is (a).

Answer:

In the given figure, we need to find the value of x.

Here, DBA is a straight line, so using the property, “angles forming a linear pair are supplementary”, we get,

Now, applying the value of y inand

Also,

Further, applying angle sum property of the triangle

In ΔABC

Thus,

Therefore, the correct option is (d).

Answer:

In the given figure, we need to find the value of x

Here, according to the angle sum property of the triangle

In ΔABD

Also, ABC is a straight line. So, using the property, “angles forming a linear pair are supplementary”, we get,

Further, using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get

Thus,

Therefore, the correct option is (d).

Answer:

In the given figure,and

We need to find the measure of x

Here, we draw a line RS parallel to BP, i.e

Also, using the property, “two lines parallel to the same line are parallel to each other”

As,

Thus,

Now, and BA is the transversal, so using the property, “alternate interior angles are equal”

Similarly, and AC is the transversal

                              ........(2)

Adding (1) and (2), we get

Also, as

Using the property,”angles opposite to equal sides are equal”, we get

Further, using the property, “an exterior angle is equal to the sum of the two opposite interior angles”

In ΔABC

Thus,

Therefore, the correct option is (c).

Answer:

In the given figure,,,and

We need to find the value of x and y

Here, we draw a line ST parallel to AB, i.e

Also, using the property, “two lines parallel to the same line are parallel to each other”

As,

Thus,

Now, and EF is the transversal, so using the property, ”alternate interior angles are equal”, we get,

Similarly,and EF is the transversal

                          .......(2)

Adding (1) and (2), we get

Further,FPE is a straight line

Applying the property, angles forming a linear pair are supplementary

Also, applying angle sum property of the triangle

In ΔPRQ

Thus,

Therefore, the correct option is (a).

Answer:

In the given problem, the exterior angles obtained on producing the base of a triangle both ways areand. So, let us draw ΔABC and extend the base BC, such that:

Here, we need to find

Now, since BCD is a straight line, using the property, “angles forming a linear pair are supplementary”, we get

Similarly, EBS is a straight line, so we get,

Further, using angle sum property in ΔABC

Thus,

Therefore, the correct option is (c).

Answer:

In the given problem, bisectors of the acute angles of a right angled triangle meet at O. We need to find .

Now, using the angle sum property of a triangle

In ΔABC

Now, further multiplying each of the term by in (1)

Also, applying angle sum property of a triangle

In ΔAOC

Thus,

Therefore, the correct option is (c).

Answer:

In the given figure, bisects of exterior anglesand meet at O and

We need to find

Now, according to the theorem, “if the sides AB and AC of a ΔABC are produced to P and Q respectively and the bisectors of and intersect at O, therefore, we get,

Hence, in ΔABC

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, bisectors of and meet at E and

We need to find

Here, using the property, “an exterior angle of the triangle is equal to the sum of the opposite interior angles”, we get,

In ΔABC with as its exterior angle

         ........(1)

Similarly, in with as its exterior angle

(CE and BE are the bisectors of and)

    .......(2)

Now, multiplying both sides of (1) by

We get, 

   ......(3)

From (2) and (3) we get,

Thus,

Therefore, the correct option is (a).

Answer:

In the given problem, BC of ΔABC is produced to point D. bisectors of meet side BC at L, and

Here, using the property, “exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,

In ΔABC

Now, as AL is the bisector of

Also, is the exterior angle of ΔALC

Thus,

Again, using the property, “exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,

In

Thus,

Therefore the correct option is (b).

Answer:

In the given problem,

We need to find the value of x

Here, as, using the property, “consecutive interior angles are supplementary”, we get

               ..........(1)

Further, applying angle sum property of the triangle 

In ΔABC

(using 1)

Now, AB is a straight line, so using the property, “angles forming a linear pair are supplementary”, we get,

Thus,

Therefore, the correct option is (c).

Answer:

In the given problem, we need to find the value of x.

Here, according to the corollary, “if bisectors of and of a ΔABC meet at a point O, then

In ΔRST

Further solving for x, we get,

Thus,

Therefore, the correct option is (d).

Answer:

Let the measure of three angles of the triangle be 5x, 3x and 7x.

Now,

$5x+3x+7x=180°$            (Angle sum property of triangle)

$\Rightarrow 15x=180°$

$\Rightarrow x=\frac{180°}{15}=12°$

$\therefore 5x=5\times 12°=60°, 3x=3\times 12°=36°$ and $7x=7\times 12°=84°$

So, the measure of the angles of the triangle are 60º, 36º and 84º.

A triangle, each of whose angles is acute, is called an acute triangle. Thus, the triangle is an acute triangle.

The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is a/an __acute__ triangle. 

 

Answer:

Let the measure of three angles of the triangle be 2x, 4x and 3x.

Now,

$2x+4x+3x=180°$            (Angle sum property of triangle)

$\Rightarrow 9x=180°$

$\Rightarrow x=\frac{180°}{9}=20°$

$\therefore 2x=2\times 20°=40°, 4x=4\times 20°=80°$ and $3x=3\times 20°=60°$

So, the measure of the angles of the triangle are 40º, 80º and 60º.

Thus, the measure of the smallest angle of the triangle is 40º.

Angles of a triangle are in the ratio 2 : 4 : 3. The measure of the smallest angle of the triangle is ___40º___.




 

Answer:

We know that the sum of the angles of a triangle is 180º.

The given angles are 53°, 64° and 63°.

Sum of the given angles = 53° + 64° + 63° = 180º

Here, the sum of the angles of the triangle is 180º. But, the measure of sides of the triangles is not known. So, infinitely many triangles can be drawn and sum of the angles of every triangle is 180º.

The number of triangles that can be drawn the measure of whose angles are 53°, 64° and 63°, is __infinite many__.

Answer:

In ∆ABC, ∠A = 130º.

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2} .....\left(1\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2} .....\left(2\right)$

Now,

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$              (Angle sum property of triangle)

$\Rightarrow 130°+\angle \mathrm{B}+\angle \mathrm{C}=180°$

$\Rightarrow \angle \mathrm{B}+\angle \mathrm{C}=180°-130°=50°$         .....(3)

In ∆BOC,

∠OBC + ∠OCB + ∠BOC = 180º             (Angle sum property of triangle)

$$\begin{gathered} \Rightarrow \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(1\right) \mathrm{and} \left(2\right)\right] \\ \Rightarrow \frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \\ \Rightarrow \frac{50°}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(3\right)\right] \\ \Rightarrow \angle \mathrm{BOC}=180°-25°=155° \end{gathered}$$

If the measure of one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles is ___155º___.

Answer:

Let the measure of each of the two interior opposite angles be x.

Measure of exterior angle of triangle = 105°

We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.

$\therefore x+x=105°$

$\Rightarrow 2x=105°$

$\Rightarrow x=\frac{105°}{2}=52.5°$

An exterior angle of a triangle is 105° and its two interior opposite angles are equal. The measure of each of these two angles is ___52.5º___.

Answer:

Let the three angles of the triangle be x, y and z.

It is given that, 

x < 60º,  y < 60º and z < 60º

Now, 

x + y + z < 60º + 60º + 60º

⇒ x + y + z < 180º

Or Sum of the angles of the triangle < 180º

This is not possible as the sum of the angles of a triangle is always equal to 180º.

Thus, no triangle can be drawn with the measure of each angle less than 60°.

The number of triangles that can be drawn with the measure of each angle less than 60°, is ____0____.

Answer:

An angle whose measure is more than 90º but less than 180º is called an obtuse angle.

If two angles of a triangle are obtuse, then the sum of these two obtuse angles would be more than 180º.

So, the sum of the three angles of the triangle in this case would be more than 180º. This is not possible as the sum of three angles of a triangle is always equal to 180º.

Thus, a triangle cannot have two obtuse angles.

A triangle cannot have two ___obtuse ___ angles.



 

Answer:

An angle whose measure is less than 90º is called an acute angle. A triangle can have three acute angles such that their sum is 180º. A triangle which has all acute angles is called an acute triangle. 

All angles of a triangle can be __acute__ angles.

Answer:

In ∆ABC, ∠A < ∠B < 45°. This means that the measure of ∠A is less than 45º and measure of ∠B is less than 45º. Also, the measure of ∠A is less than measure of ∠B.

So, ∠A < 45° and ∠B < 45°

∴ ∠A + ∠B < 45° + 45°

⇒ ∠A + ∠B < 90°

Adding ∠C to both sides, we have
 
∠A + ∠B + ∠C < 90° + ∠C       .....(1)

We know

∠A + ∠B + ∠C = 180°           (Angle sum property of triangle)

So,

180° < 90° + ∠C           [Using (1)]

Or 90° + ∠C > 180°

⇒ ∠C > 180° − 90°

⇒ ∠C > 90°

A triangle with one angle an obtuse angle is known as an obtuse triangle. So, ∆ABC is an obtuse triangle.

In a ∆ABC, if ∠A < ∠B < 45°, then ∆ABC is a/an __obtuse__ triangle.
 

Answer:

It is given that, in ∆ABC, ∠A > ∠B > ∠C and the measures of ∠A, ∠B and ∠C in degrees are integers. 

So, the least value of ∠C is 1º and ​the least value of ∠B is 2º.

In ∆ABC,

∠A + ∠B + ∠C = 180°           (Angle sum property of triangle)

∴ ∠A + 2° + 1° = 180°

⇒ ∠A = 180° − 3° = 177°

In a triangle ABC, if ∠A > ∠B > ∠C and the measures of ∠A, ∠B and ∠C in degrees are integers, then the least possible values of A, B and C are __177°,  2°__ and __1°_ respectively.


Note: The value of ∠A depends upon the values of ∠B and ∠C. The least value of ∠A would be 61º. But, it that case the values of ∠B or ∠C would not be least. 


 

Answer:

Let the measure of three angles of the triangle be x, 2x and 3x.

Now,

$x+2x+3x=180°$            (Angle sum property of triangle)

$\Rightarrow 6x=180°$

$\Rightarrow x=\frac{180°}{6}=30°$

$\therefore 2x=2\times 30°=60°$ and $3x=3\times 30°=90°$

So, the measure of the angles of the triangle are 30º, 60º and 90º.

Now, a triangle with one angle a right angle is called a right triangle. Thus, the triangle is a right triangle. 

The measures of three angles of a triangle are in the ratio 1 : 2 : 3. Then, the triangle is a ___right___ triangle.

Answer:

In ∆ABC, ∠B + ∠C = 100°.

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2} .....\left(1\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2} .....\left(2\right)$

In ∆BOC,

∠OBC + ∠OCB + ∠BOC = 180º             (Angle sum property of triangle)

$$\begin{gathered} \Rightarrow \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(1\right) \mathrm{and} \left(2\right)\right] \\ \Rightarrow \frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \\ \Rightarrow \frac{100°}{2}+\angle \mathrm{BOC}=180° \left(\mathrm{Given}\right) \\ \Rightarrow \angle \mathrm{BOC}=180°-50°=130° \end{gathered}$$


The internal bisectors of ∠B and ∠C of ∆ABC meet at O. If B + C = 100°, then ∠BOC = ____130º____.

Answer:

A plane figure bounded by three lines in a plane is called a triangle. A triangle has three sides, three angles and three vertices. The figure below represents a ΔABC, with AB, BC and CA as the three sides; ∠A, ∠B and ∠C as the three angles; A, B and C as the three vertices. 

Answer:

In the given problem, ΔABC is an obtuse triangle, withas the obtuse angle.

So, according to “the angle sum property of the triangle”, for any kind of triangle, the sum of its angles is 180°. So,

Therefore, sum of the angles of an obtuse triangle is.

Answer:

In ΔABC,,and the bisectors of and meet at O.

We need to find the measure of

Since,BO is the bisector of

Similarly,CO is the bisector of

Now, applying angle sum property of the triangle, in ΔBOC, we get,

Therefore,.

Answer:

In the given problem, angles of ΔABC are in the ratio 2:1:3

We need to find the measure of the smallest angle.

Let,

According to the angle sum property of the triangle, in ΔABC, we get,

Thus, 

Since, the measure of is the smallest of all the three angles.

Therefore, the measure of the smallest angle is .

Answer:

Exterior angle theorem states that, if a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Thus, in ΔABC

Answer:

In the given problem, the sum of two angles of a triangle is equal to its third angle.

We need to find the measure of the third angle.

Thus, it is given, in

                                ........(1)

Now, according to the angle sum property of the triangle, we get,

(Using 1)

Therefore, the measure of the third angle is.

Answer:

In the given figure,,,and

We need to find

Here, GF and CD are straight lines intersecting at point H, so using the property, “vertically opposite angles are equal”, we get,

Further, asand AC is the transversal

Using the property, “alternate interior angles are equal”

Further applying angle sum property of the triangle

In ΔGHC

Hence, applying the property, “angles forming a linear pair are supplementary”

As AGC is a straight line

Therefore,

Answer:

In the given figure,,,and

We need to find the value of x and y

Here, asand BD is the transversal, so according to the property, “alternate interior angles are equal”, we get

Similarly, as and DF is the transversal

(Using 1)

Further, EGH is a straight line. So, using the property, angles forming a linear pair are supplementary

Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,

In with as its exterior angle

Thus,

Answer:

In the given ΔABC, 

,and satisfy the relation

We need to fine the measure of.

As,

         ........(1)

Now, using the angle sum property of the triangle, we get,

(Using 1)

Therefore,

Answer:

In the given ΔABC,, the bisectors of and meet at O and

We need to find the measure of

So here, using the corollary, “if the bisectors of and of a meet at a point O, then”

Thus, in ΔABC

Thus,

Answer:

In the given problem, we need to find the difference between the sum of the exterior angles and.

Now, according to the exterior angle theorem

              .........(1)

Also,

             .........(2)

Further, adding (1) and (2)

             .........(3)

Also, according to the angle sum property of the triangle, we get,

       .........(4)

Now, we need to find the difference between the sum of the exterior angles and.

Thus, 

(Using 4)

Therefore,

Answer:

In the given ,and AB is produced to D such that

We need to find

Now, using the property, “angles opposite to equal sides are equal”

As

                ........(1)

Similarly,

As

                 ........(2)

Also, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angle” 

In ΔBDC

(Using 2)

From (1), we get

               .......(3)

Now, we need to find

That is, 

(Using 3)

(Using 2)

Eliminating from both the sides, we get 3:1 

Thus, the ratio of is

Answer:

In the given figure, bisectors of and meet at E and

We need to find

Here, using the property: an exterior angle of the triangle is equal to the sum of the opposite interior angles.

In ΔABC with as its exterior angle

    ........(1)

Similarly, in ΔBE with as its exterior angle

(CE and BE are the bisectors of and)

   ........(2)

Now, multiplying both sides of (1) by

We get, 

         ........(3)

From (2) and (3) we get,

Thus,