RD Sharma Solutions Class 7 Properties Of Triangles Exercise 15.2

RD Sharma Class 7 Solutions Chapter 15 Ex 15.2 PDF Free Download

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 \(\angle FAB+\angle ABC+\angle BCD+\angle CDE+\angle DEF+\angle EFA\)= \(720^{\circ}\)

1

Q15.Find x,y,z(whichever is required) from the figures given below

2

(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 \(\angle B\)

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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

4

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) \(\angle BOC\)

5

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.

6

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 \(\angle E\)  .

7

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 \(\angle LAD\) 

8

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 \(\angle BOD\).  

9

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 \(\angle P\) 

10

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,

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

\(\angle QPR=90^{\circ}\)

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