3

The figure explains it best.

We have $ABC$ isosceles triangle. We know a few angles as follows:

$$\begin{align} ACB &= 20°\\ PAB &= 50°\\ ABQ &= 60° \end{align}$$

Find $\angle BQP$

grg
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Gabor Magyar
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  • This is quite a popular question. What have you tried so far? – Calvin Lin Dec 15 '13 at 19:17
  • Quite a few things actually, includong just calculating the angles one after the other, and trying to apply the inscribed angle theorem... neither of them worked. – Gabor Magyar Dec 15 '13 at 19:35
  • If you don't care about simplifying your answer, then you can use the fact that triangle ABP is isoceles to compute BP, and you can also compute BQ, both using the law of sines. Then apply the law of sines once more to compute the ratio of sin BQP to sin BPQ... – Ashvin Swaminathan Dec 30 '16 at 17:46
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    Search the site (or the web) for "adventitious angles". – Gerry Myerson Dec 30 '16 at 23:58

1 Answers1

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Let us perform a preliminary angle chasing:

enter image description here

revealing that $ABP$ is an isosceles triangle. We may assume $AB=1$ and apply the sine theorem, leading to: $\frac{PR}{\sin 20^\circ}=\frac{1}{\sin 70^\circ}$, $\frac{AQ}{\sin 60^\circ}=\frac{1}{\sin 40^\circ}$, $\frac{QR}{\sin 30^\circ}=\frac{AQ}{\sin 70^\circ}$. By applying the sine theorem to $PRQ$ and simplifying the resulting expression for $\sin\widehat{PQR}=\sin\widehat{PQB}$ we get $\widehat{PQB}=\color{red}{30^\circ}$.

Jack D'Aurizio
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