The circle O is an inscribed circle of △ABC. AB = AC = 13, and BC = 10. Find sin ∠ AOR

Question

The circle $O$ is an inscribed circle of $△ABC$ and points $P, Q$ and $R$ are the points of tangency of sides $BC, CA$ and $AB$ respectively. $AB = AC = 13$ and $BC = 10$. Find $\sin ∠ AOR$.

I didn't got the right answer for this one. It's supposed to be $\frac{12}{13}$ and I got $\frac{3}{2}$.

First by applying the secant and tangent theorem I got that $$AR^2 = AO\cdot AP$$

$AR = 6$ because $BP$ is half $BC$ and $BP = BR = 5$. Then $$AR = AB - BR = 13 - 5 = 8$$

Then I got that $AB^2 = 5^2 + AP^2$. Then $AP = 12$.

Then $AR^2 = AO\cdot AP\implies 8^2 = AO \cdot 12\\ AO = 64/12 = 16/3$

Then $\sin ∠ AOR = AR/AO = 8 : 16/3 = 3/2$

What am I doing wrong?

Answer 1

Your mistake is in the tangent-secant theorem. It should be $$AR^2=AD\cdot AP$$ where $D$ is point of intersection of $AO$ with the in circle.

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Correct solution:

In $\triangle APB$, $$\sin \angle BAP=\sin \angle RAO=\frac{5}{13}$$

In $\triangle AOR$, $$\sin \angle AOR=\sin(90°-\angle RAO)=\cos \angle RAO=\frac{12}{13}$$

Answer 2

Note that because this is an isosceles triangle, $A,O$ and $P$ are colinear because all three are along the line of symmetry.

Note also that by tangency, $\angle OPB$ and $\angle ORA$ are both right. This means that by $A-A$ test, $\triangle AOR$ and $\triangle ABP$ are similar and $\angle AOR$ equals $\angle ABP$.

By the pythagoras theorem, we can then find that $AP$ is $12$, meaning $\sin(AOR)=\sin(ABP)=\frac{AP}{AB}=\frac{12}{13}$

The tangent-secant theorem doesn't apply here because $O$ is the point in the middle of the circle, not nearest $A$.