 RD Sharma Solutions Class 7 Properties Of Triangles Exercise 15.2

RD Sharma Solutions Class 7 Chapter 15 Exercise 15.2

Exercise 15.2

Q1. Two angles of a triangle are of measures $150^{\circ}$

and $30^{\circ}$. Find the measure of the third angle.

Let the third angle be x

Sum of all the angles of a triangle=$180^{\circ}$

$105^{\circ}$+ $30^{\circ}$+x= $180^{\circ}$

$135^{\circ}$+x=$180^{\circ}$

x=$180^{\circ}$$135^{\circ}$

x=$45^{\circ}$

Therefore the third angle is $45^{\circ}$

Q2. One of the angles of a triangle is $130^{\circ}$, and the other two angles are equal What is the measure of each of these equal angles?

Let the second and third angle be x

Sum of all the angles of a triangle=$180^{\circ}$

$130^{\circ}$+ x+x= $180^{\circ}$

$130^{\circ}$+2x=$180^{\circ}$

2x=$180^{\circ}$$130^{\circ}$

2x=$50^{\circ}$

x=$\frac{50}{2}$

x=$25^{\circ}$

Therefore the two other angles are $25^{\circ}$ each

Q3. The three angles of a triangle are equal to one another. What is the measure of each of the angles?

Let the each angle be x

Sum of all the angles of a triangle=$180^{\circ}$

x+x+x= $180^{\circ}$

3x=$180^{\circ}$

x=$\frac{180}{3}$

x=$60^{\circ}$

Therefore angle is $60^{\circ}$ each

Q4. If the angles of a triangle are in the ratio 1 : 2 : 3, determine three angles.

If angles of the triangle are in the ratio 1:2:3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’

Sum of all the angles of a triangle=$180^{\circ}$

x+2x+3x= $180^{\circ}$

6x=$180^{\circ}$

x=$\frac{180}{6}$

x=$30^{\circ}$

2x=$30^{\circ}\times2$=$60^{\circ}$

3x=$30^{\circ}\times3$=$90^{\circ}$

Therefore the first angle is $30^{\circ}$, second angle is $60^{\circ}$ and third angle is $90^{\circ}$

Q5. The angles of a triangle are $\left(x-40 \right)^{\circ}$ , $\left(x-20 \right)^{\circ}$ and $\left(\frac{1}{2}-10 \right)^{\circ}$. Find the value of x.

Sum of all the angles of a triangle=$180^{\circ}$

$\left(x-40 \right )^{\circ}+\left(x-20 \right )^{\circ}+\left(\frac{x}{2}-10 \right )^{\circ}=180^{\circ}\\ x+x+\frac{x}{2}-40^{\circ}-20^{\circ}-10^{\circ}=180^{\circ}\\ x+x+\frac{x}{2}-70^{\circ}=180^{\circ}\\ x+x+\frac{x}{2}=180^{\circ}+70^{\circ}\\ \frac{5x}{2}=250^{\circ}\\ x=\frac{2}{5}\times 250^{\circ}\\ x=100^{\circ}$

Hence we can conclude that x is equal to $100^{\circ}$

Q6. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is $10^{\circ}$. Find the three angles.

Let the first angle be x

Second angle be x+$10^{\circ}$

Third angle be x+$10^{\circ}$+ $10^{\circ}$

Sum of all the angles of a triangle=$180^{\circ}$

x+ x+$10^{\circ}$+ x+$10^{\circ}$+ $10^{\circ}$= $180^{\circ}$

3x+30=180

3x=180-30

3x=150

x=$\frac{150}{3}$

x=$50^{\circ}$

First angle is 50

Second angle x+$10^{\circ}$=50+10= $60^{\circ}$

Third angle x+$10^{\circ}$+ $10^{\circ}$=50+10+10= $70^{\circ}$

Q7. Two angles of a triangle are equal and the third angle is greater than each of those angles by $30^{\circ}$. Determine all the angles of the triangle

Let the first and second angle be x

The third angle is greater than the first and second by $30^{\circ}$=x+$30^{\circ}$

The first and the second angles are equal

Sum of all the angles of a triangle=$180^{\circ}$

x+x+x+$30^{\circ}$= $180^{\circ}$

3x+30=180

3x=180-30

3x=150

x=$\frac{150}{3}$

x=$50^{\circ}$

Third angle=x+$30^{\circ}$= $50^{\circ}$+ $30^{\circ}$= $80^{\circ}$

The first and the second angle is $50^{\circ}$ and the third angle is $80^{\circ}$

Q8. If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.

One angle of a triangle is equal to the sum of the other two

x=y+z

Let the measure of angles be x,y,z

x+y+z=$180^{\circ}$

x+x=$180^{\circ}$

2x=$180^{\circ}$

x=$\frac{180^{\circ}}{2}$

x=$90^{\circ}$

If one angle is $90^{\circ}$ then the given triangle is a right angled triangle

Q9. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Each angle of a triangle is less than the sum of the other two

Measure of angles be x,y and z

x>y+z

y<x+z

z<x+y

Therefore triangle is an acute triangle

Q10. In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:

(i) $63^{\circ}$, $37^{\circ}$, $80^{\circ}$

(ii) $45^{\circ}$, $61^{\circ}$, $73^{\circ}$

(iii) $59^{\circ}$, $72^{\circ}$, $61^{\circ}$

(iv) $45^{\circ}$, $45^{\circ}$, $90^{\circ}$

(v) $30^{\circ}$, $20^{\circ}$, $125^{\circ}$

(i) $63^{\circ}$, $37^{\circ}$, $80^{\circ}$= $180^{\circ}$

Angles form a triangle

(ii) $45^{\circ}$, $61^{\circ}$, $73^{\circ}$ is not equal to $180^{\circ}$

Therefore not a triangle

(iii) $59^{\circ}$, $72^{\circ}$, $61^{\circ}$ is not equal to $180^{\circ}$

Therefore not a triangle

(iv) $45^{\circ}$, $45^{\circ}$, $90^{\circ}$= $180$

Angles form a triangle

(v) $30^{\circ}$, $20^{\circ}$, $125^{\circ}$ is not equal to $180^{\circ}$

Therefore not a triangle

Q11. The angles of a triangle are in the ratio 3: 4 : 5. Find the smallest angle

Given  that

Angles of a triangle are in the ratio: 3: 4: 5

Measure of the angles be 3x, 4x, 5x

Sum of  the angles of a triangle=$180^{\circ}$

3x+4x+5x=$180^{\circ}$

12x=$180^{\circ}$

x=$\frac{180^{\circ}}{12}$

x=$15^{\circ}$

Smallest angle=3x

=3 x $15^{\circ}$

=$45^{\circ}$

Q12. Two acute angles of a right triangle are equal. Find the two angles.

Given acute angles of a right angled triangle are equal

Right triangle: whose one of the angle is a right angle

Measured angle be x,x,$90^{\circ}$

x+x+$180^{\circ}$ =$180^{\circ}$

2x=$90^{\circ}$

x=$\frac{90^{\circ}}{2}$

x=$45^{\circ}$

The two angles are $45^{\circ}$  and $45^{\circ}$

Q13. One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?

Angle of a triangle is greater than the sum of the other two

Measure of the angles be x,y,z

x>y+z    or

y>x+z    or

z>x+y

x or y or z>$90^{\circ}$ which is obtuse

Therefore triangle is an obtuse angle

Q14. AC, AD and AE are joined. Find $\angle FAB+\angle ABC+\angle BCD+\angle CDE+\angle DEF+\angle EFA$

$\angle FAB+\angle ABC+\angle BCD+\angle CDE+\angle DEF+\angle EFA$

We know that sum of the angles of a triangle is $180^{\circ}$

Therefore in $\triangle ABC$,we have

$\angle CAB+\angle ABC+\angle BCA=180^{\circ}$—(i)

In $\triangle ACD$,we have

$\angle DAC+\angle ACD+\angle CDA=180^{\circ}$—(ii)

In $\triangle ADE$,we have

$\angle EAD+\angle ADE+\angle DEA=180^{\circ}$—(iii)

In $\triangle AEF$,we have

$\angle FAE+\angle AEF+\angle EFA=180^{\circ}$—(iv)

Adding (i),(ii),(iii),(iv) we get

$\angle CAB+\angle ABC+\angle BCA+\angle DAC+\angle ACD+\angle CDA+\angle EAD+\angle ADE+\angle DEA +\angle FAE+\angle AEF+\angle EFA$= $720^{\circ}$

Therefore $\angle FAB+\angle ABC+\angle BCD+\angle CDE+\angle DEF+\angle EFA$= $720^{\circ}$ Q15.Find x,y,z(whichever is required) from the figures given below (i)

In $\triangle ABC$ and $\triangle ADE$ we have :

$\angle ADE=\angle ABC$ (corresponding angles)

x=$40^{\circ}$

$\angle AED=\angle ACB$ (corresponding angles)

y=$30^{\circ}$

We know that the sum of all the three angles of a triangle is equal to $180^{\circ}$

x +y +z = $180^{\circ}$ (Angles of A ADE)

Which means : $40^{\circ}$ +$30^{\circ}$  + z = $180^{\circ}$

z= $180^{\circ}-70^{\circ}$

z=$110^{\circ}$

Therefore, we can conclude that the three angles of the given triangle are $40^{\circ}$, $30^{\circ}$  and $110^{\circ}$.

(ii) We can see that in $\triangle ADC$, $\angle ADC$  is equal to $90^{\circ}$.

($\triangle ADC$  is a right triangle)

We also know that the sum of all the angles of a triangle is equal to $180^{\circ}$.

Which means : $45^{\circ}$+ $90^{\circ}$+y=$180^{\circ}$ (Sum of the angles of $\triangle ADC$  )

$135^{\circ}$  + y = $180^{\circ}$

y = $180^{\circ}$$135^{\circ}$.

y =$45^{\circ}$.

We can also say that in $\triangle ABC$,  $\angle ABC+\angle ACB+\angle BAC$  is equal to $180^{\circ}$.

(Sum of the angles of A ABC)

$40^{\circ}$ + y + (x + $45^{\circ}$) = $180^{\circ}$

$40^{\circ}$ + $45^{\circ}$ + x + $45^{\circ}$ = $180^{\circ}$                                                 (y = $45^{\circ}$)

x= $180^{\circ}$$130^{\circ}$

x= $50^{\circ}$

Therefore, we can say that the required angles are $45^{\circ}$ and $50^{\circ}$.

(iii) We know that the sum of all the angles of a triangle is equal to $180^{\circ}$.

Therefore, for $\triangle ABD$:

$\angle ABD+\angle ADB+\angle BAD=180^{\circ}$ (Sum of the angles of $\triangle ABD$)

$50^{\circ}$ + x + $50^{\circ}$  = $180^{\circ}$

$100^{\circ}$  +x = $180^{\circ}$

x = $180^{\circ}$$100^{\circ}$

x =$80^{\circ}$

For $\triangle ABC$:

$\angle ABC+\angle ACB+\angle BAC=180^{\circ}$ (Sum of the angles of $\triangle ABC$)

$50^{\circ}$   + z + ($50^{\circ}$   + $30^{\circ}$  ) = $180^{\circ}$

$50^{\circ}$   + z + $50^{\circ}$ + $30^{\circ}$   = $180^{\circ}$

z = $180^{\circ}$$130^{\circ}$

z = $50^{\circ}$

Using the same argument for $\triangle ADC$:

$\angle ADC+\angle ACD+\angle DAC=180^{\circ}$ (Sum of the angles of $\triangle ADC$)

y +z +$30^{\circ}$ =$180^{\circ}$

y + $50^{\circ}$ + $30^{\circ}$ =$180^{\circ}$    ( z =$50^{\circ}$   )

y = $180^{\circ}$$80^{\circ}$

y = $100^{\circ}$

Therefore, we can conclude that the required angles are $80^{\circ}$, $50^{\circ}$ and $100^{\circ}$.

(iv) In $\triangle ABC$ and $\triangle ADE$ we have :

$\angle ADE=\angle ABC$ (Corresponding angles)

y = $50^{\circ}$

Also, $\angle AED=\angle ACB$  (Corresponding angles)

z = $40^{\circ}$

We know that the sum of all the three angles of a triangle is equal to $180^{\circ}$.

Which means : x+$50^{\circ}$ +$40^{\circ}$ =$180^{\circ}$  (Angles of $\triangle ADE$)

x = $180^{\circ}$$90^{\circ}$

x = $90^{\circ}$

Therefore, we can conclude that the required angles are $50^{\circ}$, $40^{\circ}$ and $90^{\circ}$.

Q16. If one angle of a triangle is $60^{\circ}$ and the other two angles are in the ratio 1 :2, find the angles

We know that one of the angles of the given triangle is $60^{\circ}$. (Given)

We also know that the other two angles of the triangle are in the ratio 1 : 2.

Let one of the other two angles be x.

Therefore, the second one will be 2x.

We know that the sum of all the three angles of a triangle is equal to $180^{\circ}$.

$60^{\circ}$ +x + 2x = $180^{\circ}$

3x =$180^{\circ}$$60^{\circ}$

3x = $120^{\circ}$

x = $\frac{120^{\circ}}{3}$

x = $40^{\circ}$

2x = 2 x 40

2x = $80^{\circ}$

Hence, we can conclude that the required angles are $40^{\circ}$ and $80^{\circ}$.

Q17. It one angle of a triangle is $100^{\circ}$ and the other two angles are in the ratio 2 : 3. find the angles.

We know that one of the angles of the given triangle is $100^{\circ}$.

We also know that the other two angles are in the ratio 2 : 3.

Let one of the other two angles be 2x.

Therefore, the second angle will be 3x.

We know that the sum of all three angles of a triangle is $180^{\circ}$.

$100^{\circ}$  + 2x + 3x = $180^{\circ}$

5x = $180^{\circ}$$100^{\circ}$

5x = $80^{\circ}$

x = $\frac{80^{\circ}}{5}$

2x = 2 x 16

2x = $32^{\circ}$

3x = 3 x 16

3x = $48^{\circ}$

Thus, the required angles are $32^{\circ}$  and $48^{\circ}$.

Q18. In $\triangle ABC$, if $3\angle A=4\angle B=6\angle C$, calculate the angles.

We know that for the given triangle, $3\angle A=6\angle C$

$\angle A=2\angle C$—(i)

We also know that for the same triangle, $4\angle B=6\angle C$

$\angle B=\frac{6}{4}\angle C$—(ii)

We know that the sum of all three angles of a triangle is $180^{\circ}$.

Therefore, we can say that:

$\angle A+\angle B+\angle C=180^{\circ}$ (Angles of $\triangle ABC$)—(iii)

On putting the values of $\angle A\;and\;\angle B$ in equation (iii), we get :

$2\angle C+\frac{6}{4}\angle C+\angle C=180^{\circ}$

$\frac{18}{4}\ angle C=180^{\circ}$

$\ angle C=40^{\circ}$

From equation (i), we have:

$\ angle A=2\angle C=2\times40$

$\ angle A=80^{\circ}$

From equation (ii), we have:

$\ angle B=\frac{6}{4}\angle C=\frac{6}{4}\times40^{\circ}$

$\ angle B=60^{\circ}$

$\ angle A=80^{\circ}$, $\ angle B=60^{\circ}$, $\ angle C=40^{\circ}$

Therefore, the three angles of the given triangle are $80^{\circ}$, $60^{\circ}$, and $40^{\circ}$.

Q19. Is it possible to have a triangle, in which

(i) Two of the angles are right?

(ii) Two of the angles are obtuse?

(iii) Two of the angles are acute?

(iv) Each angle is less than $60^{\circ}$?

(v) Each angle is greater than $60^{\circ}$?

(vi) Each angle is equal to $60^{\circ}$

Give reasons in support of your answer in each case.

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.

(ii) No, because as we know that the sum of all three angles of a triangle is always $180^{\circ}$. If there are two obtuse angles, then their sum will be more than $180^{\circ}$, which is not possible in case of a triangle.

(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.

(iv) No, because if each angle is less than $60^{\circ}$, then the sum of all three angles will be less than $180^{\circ}$, which is not possible in case of a triangle.

Proof:

Let the three angles of the triangle be $\angle A$, $\angle B$  and $\angle C$.

As per the given information,

$\angle A$ < $60^{\circ}$  … (i)

$\angle B$< $60^{\circ}$   …(ii)

$\angle C$< $60^{\circ}$   … (iii)

On adding (i), (ii) and (iii), we get :

$\angle A$ + $\angle B$ + $\angle C$ < $60^{\circ}$+ $60^{\circ}$+ $60^{\circ}$

$\angle A$ + $\angle B$ + $\angle C$ < $180^{\circ}$

We can see that the sum of all three angles is less than $180^{\circ}$, which is not possible for a triangle.

Hence, we can say that it is not possible for each angle of a triangle to be less than $60^{\circ}$.

(v) No, because if each angle is greater than $60^{\circ}$, then the sum of all three angles will be greater than $180^{\circ}$, which is not possible.

Proof:

Let the three angles of the triangle be $\angle A$, $\angle B$  and $\angle C$. As per the given information,

$\angle A$ > $60^{\circ}$  … (i)

$\angle B$>$60^{\circ}$   …(ii)

$\angle C$> $60^{\circ}$   … (iii)

On adding (i), (ii) and (iii), we get:

$\angle A$ + $\angle B$ + $\angle C$ > $60^{\circ}$+ $60^{\circ}$+ $60^{\circ}$

$\angle A$ + $\angle B$ + $\angle C$ > $180^{\circ}$

We can see that the sum of all three angles of the given triangle are greater than $180^{\circ}$, which is not possible for a triangle.

Hence, we can say that it is not possible for each angle of a triangle to be greater than $60^{\circ}$.

(vi) Yes, if each angle of the triangle is equal to $60^{\circ}$ , then the sum of all three angles will be $180^{\circ}$ , which is possible in case of a triangle.

Proof:

Let the three angles of the triangle be $\angle A$, $\angle B$  and $\angle C$. As per the given information,

$\angle A$ = $60^{\circ}$  … (i)

$\angle B$=$60^{\circ}$   …(ii)

$\angle C$= $60^{\circ}$   … (iii)

On adding (i), (ii) and (iii), we get:

$\angle A$ + $\angle B$ + $\angle C$ = $60^{\circ}$+ $60^{\circ}$+ $60^{\circ}$

$\angle A$ + $\angle B$ + $\angle C$ =$180^{\circ}$

We can see that the sum of all three angles of the given triangle is equal to $180^{\circ}$ , which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to $60^{\circ}$ .

Q20. In $\triangle ABC$, $\angle A=100^{\circ}$, AD bisects $\angle A$ and AD perpendicular BC. Find $\angle B$ Consider $\triangle ABD$

$\angle BAD$ = $\frac{100}{2}$                  (AD bisects $\angle A$)

$\angle BAD$ = $50^{\circ}$

$\angle ADB$ = $90^{\circ}$                                     (AD perpendicular to BC)

We know that the sum of all three angles of a triangle is $180^{\circ}$.

Thus,

$\angle ABD+\angle BAD+\angle ADB$= $180^{\circ}$  (Sum of angles of $\triangle ABD$)

Or,

$\angle ABD$ + $50^{\circ}$ + $90^{\circ}$ =$180^{\circ}$

$\angle ABD$ =$180^{\circ}$– $140^{\circ}$

$\angle ABD$= $40^{\circ}$

Q21. In $\triangle ABC$$\angle A=50^{\circ}$, $\angle B=100^{\circ}$  and bisector of $\angle C$  meets AB in D. Find the angles of the triangles ADC and BDC We know that the sum of all three angles of a triangle is equal to $180^{\circ}$.

Therefore, for the given $\triangle ABC$, we can say that :

$\angle A$ + $\angle B$ + $\angle C$ = $180^{\circ}$ (Sum of angles of $\triangle ABC$)

$50^{\circ}$ + $70^{\circ}$ + $\angle C$  = $180^{\circ}$

$\angle C$= $180^{\circ}$$120^{\circ}$

$\angle C$ = $60^{\circ}$

$\angle ACD$ = $\angle BCD$ =$\frac{\angle C}{2}$ (CD bisects $\angle C$  and meets AB in D. )

$\angle ACD$  = $\angle BCD$  = $\frac{60^{\circ}}{2}$= $30^{\circ}$

Using the same logic for the given $\triangle ACD$, we can say that :

$\angle DAC$  + $\angle ACD$  + $\angle ADC$  = $180^{\circ}$

$50^{\circ}$ +$30^{\circ}$ +$\angle ADC$= $180^{\circ}$

$\angle ADC$ = $180^{\circ}$$80^{\circ}$

$\angle ADC$= $100^{\circ}$

If we use the same logic for the given $\triangle BCD$, we can say that

$\angle DBC$   +$\angle BCD$   +$\angle BDC$   = $180^{\circ}$

$70^{\circ}$ + $30^{\circ}$ + $\angle BDC$   = $180^{\circ}$

$\angle BDC$   = $180^{\circ}$$100^{\circ}$

$\angle BDC$   = $80^{\circ}$

Thus,

For $\triangle ADC$: $\angle A=50^{\circ}$, $\angle D=100^{\circ}$   $\angle C=30^{\circ}$

$\triangle BDC$: $\angle B=70^{\circ}$, $\angle D=80^{\circ}$   $\angle C=30^{\circ}$

Q22. In $\triangle ABC$, $\angle A=60^{\circ}$, $\angle B=80^{\circ}$,  and the bisectors of $\angle B$  and $\angle C$,  meet at O. Find

(i) $\angle C$

(ii) $\angle BOC$ We know that the sum of all three angles of a triangle is $180^{\circ}$.

Hence, for $\triangle ABC$, we can say that :

$\angle A$ + $\angle B$ + $\angle C$ = $180^{\circ}$ (Sum of angles of $\triangle ABC$)

$60^{\circ}$+ $80^{\circ}$ + $\angle C$= $180^{\circ}$.

$\angle C$= $180^{\circ}$$140^{\circ}$

$\angle C$= $140^{\circ}$.

For $\triangle OBC$,

$\angle OBC$ =$\frac{\angle B}{2}=\frac{80^{\circ}}{2}$ (OB bisects $\angle B$ )

$\angle OBC=40^{\circ}$

$\angle OCB$ =$\frac{\angle C}{2}=\frac{40^{\circ}}{2}$ (OC bisects $\angle C$ )

$\angle OCB=20^{\circ}$

If we apply the above logic to this triangle, we can say that :

$\angle OCB$ + $\angle OBC$ + $\angle BOC$ = $180^{\circ}$ (Sum of angles of $\triangle OBC$)

$20^{\circ}$+ $40^{\circ}$  +$\angle BOC$  = $180^{\circ}$

$\angle BOC$= $180^{\circ}$$60^{\circ}$

$\angle BOC$= $120^{\circ}$

Q23. The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors. We know that the sum of all three angles of a triangle is $180^{\circ}$.

Hence, for $\triangle ABC$ , we can say that :

$\angle A$ + $\angle B$ + $\angle C$ = $180^{\circ}$

$\angle A$    + $90^{\circ}$    + $\angle C$    = $180^{\circ}$

$\angle A$   + $\angle C$    = $180^{\circ}$$90^{\circ}$

$\angle A$   + $\angle C$    = $90^{\circ}$

For $\triangle OAC$   :

$\angle OAC$   = $\frac{\angle A}{2}$                                     (OA bisects LA)

$\angle OCA$   = $\frac{\angle C}{2}$                                      (OC bisects LC)

On applying the above logic to $\triangle OAC$, we get :

$\angle AOC$+ $\angle OAC$+$\angle OCA$   = $180^{\circ}$    (Sum of angles of $\triangle AOC$   )

$\angle AOC$+ $\frac{\angle A}{2}$+ $\frac{\angle C}{2}$= $180^{\circ}$

$\angle AOC$+ $\frac{\angle A+\angle C}{2}$= $180^{\circ}$

$\angle AOC$+ $\frac{90^{\circ} }{2}$= $180^{\circ}$

$\angle AOC$= $180^{\circ}$$45^{\circ}$

$\angle AOC$= $135^{\circ}$

Q24. In $\triangle ABC$, $\angle A=50^{\circ}$ and BC is produced to a point D. The bisectors of $\angle ABC$  and $\angle ACD$ meet at E. Find $\angle E$  . In the given triangle,

$\angle ACD$= $\angle A$ + $\angle B$   . (Exterior angle is equal to the sum of two opposite interior angles.)

We know that the sum of all three angles of a triangle is $180^{\circ}$   .

Therefore, for the given triangle, we can say that :

$\angle ABC$+ $\angle BCA$ + $\angle CAB$ = $180^{\circ}$ (Sum of all angles of $\triangle ABC$  )

$\angle A$ + $\angle B$ + $\angle BCA$= $180^{\circ}$

$\angle BCA$=180°- ($\angle A$ + $\angle B$   )

$\angle ECA$= $\frac{\angle ACD}{2}$         ( EC bisects $\angle ACD$   )

$\angle ECA$= $\frac{\angle A+\angle B}{2}$        ($\angle ACD=\angle A+\angle B$   )

$\angle EBC$= $\frac{\angle ABC}{2}=\frac{\angle B}{2}(EB bisects\angle ABC)$

$\angle ECB$= $\angle ECA$ + $\angle BCA$

$\angle ECB$= $\frac{\angle A+\angle B}{2}+180^{\circ}-(\angle A+\angle B)$

If we use the same logic for $\triangle EBC$  , we can say that :

$\angle EBC$+ $\angle ECB$ + $\angle BEC$ = $180^{\circ}$ (Sum of all angles of $\triangle EBC$  )

$\frac{\angle B}{2}+\frac{\angle A+\angle B}{2}+180^{\circ}-(\angle A+\angle B)+\angle BEC=180^{\circ}$

$\angle BEC= \angle A+\angle B-(\frac{\angle A+\angle B}{2}-\frac{\angle B}{2}$

$\angle BEC=\frac{\angle A}{2}$

$\angle BEC=\frac{50^{\circ}}{2}=25^{\circ}$

Q25. In $\triangle ABC$, $\angle B=60^{\circ}$, $\angle C=40^{\circ}$, AL  perpendicular BC and AD bisects $\angle A$  such that L and D lie on side BC. Find $\angle LAD$ We know that the sum of all angles of a triangle is $180^{\circ}$

Therefore, for $\triangle ABC$, we can say that :

$\angle A$ + $\angle B$ + $\angle C$ = $180^{\circ}$

Or,

$\angle A$ + $60^{\circ}$ + $40^{\circ}$ = $180^{\circ}$

$\angle A$= $80^{\circ}$

$\angle DAC$ = $\frac{\angle A}{2}$    (AD bisects $\angle A$)

$\angle DAC$ = $\frac{80^{\circ}}{2}$

If we use the above logic on $\triangle ADC$, we can say that :

$\angle ADC$ + $\angle DCA$ + $\angle DAC$ = $180^{\circ}$ (Sum of all the angles of $\triangle ADC$)

$\angle ADC$ + $40^{\circ}$ + $40^{\circ}$ = $180^{\circ}$

$\angle ADC$=$180^{\circ}$ + $80^{\circ}$

$\angle ADC$=$\angle ALD$ + $\angle LAD$(Exterior angle is equal to the sum of two Interior opposite angles.)

$100^{\circ}$ = $90^{\circ}$+ $\angle LAD$       (AL perpendicular toBC)

$\angle LAD=90^{\circ}$

Q26. Line segments AB and CD intersect at O such that AC perpendicular DB. It $\angle CAB=35^{\circ}$ and $\angle CDB=55^{\circ}$ . Find $\angle BOD$. We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.

$\angle CAB$= $\angle DBA$ (Alternate interior angles)

$\angle DBA=35^{\circ}$

We also know that the sum of all three angles of a triangle is $180^{\circ}$.

Hence, for $\triangle OBD$, we can say that :

$\angle DBO$ + $\angle ODB$+ $\angle BOD$= $180^{\circ}$

$35^{\circ}$+ $55^{\circ}$+ $\angle BOD$ = $180^{\circ}$ ($\angle DBO=\angle DBA$ and $\angle ODB=\angle CDB$)

$\angle BOD$ =$180^{\circ}-90^{\circ}$

$\angle BOD$ = $90^{\circ}$

Q27. In Fig. 22, $\triangle ABC$   is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find $\angle P$ In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

$\angle QCA$ = $\angle CQP$ (Alternate interior angles)

Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,

$\angle ABC$ = $\angle PRQ$ (alternate interior angles).

We know that the sum of all three angles of a triangle is $180^{\circ}$.

Hence, for $\triangle ABC$, we can say that :

$\angle ABC$ + $\angle ACB$ + $\angle BAC$ = $180^{\circ}$

$\angle ABC$ + $\angle ACB$ + $90^{\circ}$ = $180^{\circ}$ (Right angled at A)

$\angle ABC$ + $\angle ACB$ = $90^{\circ}$

Using the same logic for $\triangle PQR$, we can say that :

$\angle PQR$ + $\angle PRQ$ + $\angle QPR$ = $180^{\circ}$

$\angle ABC$ + $\angle ACB$ + $\angle QPR$ =$180^{\circ}$ ($\angle ABC$ =$\angle PRQ$ and $\angle QCA$ =$\angle CQP$ )

Or,

$90^{\circ}$+ $\angle QPR$ =$180^{\circ}$ ($\angle ABC$+ $\angle ACB$ = $90^{\circ}$ )

$\angle QPR=90^{\circ}$