If ΔBPQ ≅ ΔASR and ∠A = 1/3 ∠R = ∠S then find ∠Q. (All angles are in…

∠A = \(\frac{1}{3}\)∠R

⇒ 3∠A = ∠R

Again,

\(\frac{1}{3}\)∠R = ∠S

⇒ ∠R = 3∠S

Also, ∠A = ∠S

Now,

∠A + ∠R + ∠S = \({180}^\circ\)

⇒ ∠A + 3∠A + ∠A = \({180}^\circ\)

⇒ 5∠A = \({180}^\circ\)

⇒ ∠A = \({36}^\circ\)

So, ∠R = 3 x \({36}^\circ\)

⇒ \({108}^\circ\)

Now,

As, ΔBPQ ≅ ΔASR

So, ∠R = ∠Q

So, ∠Q = \({108}^\circ\)

Therefore, the answer is \({108}^\circ\)