Practicing Success
If ΔBPQ ≅ ΔASR and ∠A = $\frac{1}{3}$ ∠R = ∠S then find ∠Q. (All angles are in degrees). |
108o 36o 72o 118o |
108o |
∠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 \({108}^\circ\) |