In Fig. 6.1, side QR of a ∆PQR has been produced to the point S. If ∠PRS = 115° and ∠P = 45°, then ∠Q is equal to
a. 70°
b. 105°
c. 51°
d. 80°

Solution:

Given, the side QR of a ∆PQR has been produced to the point S.

∠PRS = 115° and ∠P = 45°

We have to find the value of ∠Q.

We know that the sum of a linear pair of angles is equal to 180 degrees.

So, ∠R + 115° = 180°

∠R = 180° - 115°

∠R = 65°

We know that the sum of all three interior angles of a triangle is always equal to 180 degrees.

So, ∠P + ∠Q + ∠R = 180°

45° + ∠Q + 65° = 180°

110° + ∠Q = 180°

∠Q = 180° - 110°

Therefore, ∠Q = 70°

✦ Try This: In the given figure, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD

In the given figure, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 6

NCERT Exemplar Class 7 Maths Chapter 6 Sample Problem 1

In Fig. 6.1, side QR of a ∆PQR has been produced to the point S. If ∠PRS = 115° and ∠P = 45°, then ∠Q is equal to, a. 70°, b. 105°, c. 51°, d. 80°

Summary:

In Fig. 6.1, side QR of a ∆PQR has been produced to the point S. If ∠PRS = 115° and ∠P = 45°, then ∠Q is equal to 70°

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