Track / Stage 4 / 229 of 340 #489 of 1964

Problem 489

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

6. Given in $\triangle A B C$ , $\angle A, \angle B$ are acute angles, and $\sin A$ $=\frac{5}{13}, \tan B=2, A B=29 \mathrm{~cm}$ . Then the area of $\triangle A B C$ is $\qquad$ $\mathrm{cm}^{2}$

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

6.145 .

Draw a perpendicular from point $C$ to $AB$ . Let the foot of the perpendicular be $D$ .
$\because \sin A=\frac{5}{13}=\frac{CD}{AC}$ , let $m>0$ ,
$\therefore CD=5m, AC=13m$ .
$\because \tan B=\frac{CD}{BD}=2$ , we can set $n>0, CD=2n, BD=n$ ,
$\therefore BD=n=\frac{CD}{2}=\frac{5}{2}m$ .
$\therefore AD=\sqrt{(13m)^2-(5m)^2}=12m$ .
Thus, $AB=AD+BD=12m+\frac{5}{2}m=\frac{29}{2}m$ .
From $29=\frac{29}{2}m$ , we get $m=2$ . Then $CD=5m=10$ .
Therefore, $S_{\triangle ABC}=\frac{1}{2} AB \cdot CD=\frac{1}{2} \times 29 \times 10=145(\text{cm}^2)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.