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Set 9 Question 29

In a $\triangle ABC, \angle A= 54^\circ, \angle C= 24^\circ, P$ is a point on $AC$ such that $AP=BC.$ Find $\angle CBP$

I tried this problem by first finding the $\angle B$ as $102^\circ,$ then by substituting some angles as $x$ or $180^\circ-x$ and eventually I was able to make most angles in the triangle as a value of $x$ but eventually the $x$'s started cancelling out and it was making no sense and led me to nowhere.

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    can you use trigonometric functions? – 2017-01-08

1 Answers 1

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By Sine Formula,

$$\frac{\sin54^{\circ}}{BP}=\frac{\sin\alpha}{AP}$$ $$\frac{\sin24^{\circ}}{BP}=\frac{\sin\beta}{BC}$$

Since $AP=BC$, divide both equations we get

$$\frac{\sin54^{\circ}}{\sin24^{\circ}}=\frac{\sin\alpha}{\sin\beta}=\frac{\sin(\beta-54^{\circ})}{\sin\beta}$$ $$\sin54^{\circ}\sin\beta=\sin24^{\circ}\sin(\beta-54^{\circ})$$ $$\sin54^{\circ}\sin\beta=\sin24^{\circ}[\sin\beta\cos54^{\circ}-\sin54^{\circ}\cos\beta]$$ $$\sin54^{\circ}=\sin24^{\circ}[\cos54^{\circ}-\sin54^{\circ}\cot\beta]$$

Solve for $\beta$ and thus $\angle PBC$. enter image description here

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    How did you get Sin54/Sin24 = SinA/ SinB = Sin(B-54)/SinB?... I don't get the last part – 2017-01-09
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    In a triangle, the sum of the 2 interior angles is the corresponding external angle, thus $\alpha+54^{\circ}=\beta$ – 2017-01-10