In Fig. 5.28, PQ || RS. If ∠1=(2a+b)° and ∠6=(3a-b)°, then the measure of ∠2 in terms of b is
a. (2 + b)°
b. (3 - b)°
c. (108 - b)°
d. (180 - b)°
Solution:
Given, PQ || RS
Also, ∠1 = (2a + b)° and ∠6 = (3a - b)°
We have to find the measure of ∠2 in terms of b.
If two parallel lines are intersected by a transversal, each pair of corresponding angles is equal.
From the figure,
Corresponding angles, ∠1 = ∠5
So, ∠5 = (2a + b)°
We know that the sum of a linear pair of angles is always equal to 180 degrees.
So, ∠5 + ∠6 = 180°
(2a + b)° + (3a - b)° = 180°
On rearranging,
2a° + 3a° + b° - b° = 180°
5a° = 180°
a = 180°/5
a = 36°
Similarly, ∠1 + ∠2 = 180°
(2a + b)° + ∠2 = 180°
2a° + b° + ∠2 = 180°
2(36°) + b° + ∠2 = 180°
72° + b° + ∠2 = 180°
∠2 = 180° - 72° - b°
∠2 = 108° - b°
Therefore, ∠2 = (108 - b)°
✦ Try This: In Fig, ∠1 = 60° and ∠2 = ( 2/3 )rd of a right angle. Prove that l || m.
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 5
NCERT Exemplar Class 7 Maths Chapter 5 Problem 35
In Fig. 5.28, PQ || RS. If ∠1=(2a+b)° and ∠6=(3a-b)°, then the measure of ∠2 in terms of b is: a. (2+b)°, b. (3-b)°, c. (108-b)°, d. (180-b)°
Summary:
In Fig. 5.28, PQ || RS. If ∠1=(2a+b)° and ∠6=(3a-b)°, then the measure of ∠2 in terms of b is (108 - b)°
☛ Related Questions:
- In Fig. 5.29, PQ||RS and a : b = 3 : 2. Then, f is equal to: a. 36°, b. 108°, c. 72°, d. 144°
- In Fig. 5.30, line l intersects two parallel lines PQ and RS. Then, which one of the following is no . . . .
- In Fig. 5.30, which one of the following is not true? a. ∠1 + ∠5 = 180°, b. ∠2 + ∠5 = 180°, c. ∠3 + . . . .