Trigonometry
Properties of Triangles
MMTS_Full_Test_08
Grade 12
Question:
In $\triangle ABC$, $\angle A = \tan^{-1}7$, $\angle C = \tan^{-1}\frac{4}{3}$. Let $D$ be an interior point on side $AC$ such that area of $\triangle ABD$ is twice the area of $\triangle BCD$. If $\angle ABD = \theta$, then the value of $\tan 2\theta$ is
$\frac{12}{5}$
$\frac{5}{12}$
$\frac{36}{77}$
$\frac{77}{36}$
Step-by-Step Solution
Key Concept: Area ratio $= AD/DC = 2$. Use cevian angle formula: $\cot\theta = 2\cot(\angle ABD) - \cot(\angle ABC)$... use $\angle ABD=\theta$, $\angle DBC=\angle B-\theta$.
$\tan A=7$, $\tan C=4/3$, $\tan B=-\tan(A+C)$. $\tan(A+C)=\frac{7+4/3}{1-28/3}=\frac{25/3}{-25/3}=-1$. So $B=3\pi/4$. $\tan B=-1$... wait $\tan(A+C)=-1$ means $A+C=3\pi/4$, $B=\pi/4$, $\tan B=1$. Area ratio $AD/DC=2$; by angle bisector-like formula with ratio $2:1$: $\cot\theta - \cot(B-\theta)=3\cot B-\cot B=...$; using $AD/DC=2$: $\frac{\sin(B-\theta)}{\sin\theta}=\frac{DC}{AD}\cdot\frac{AB}{BC}\cdot\frac{\sin A}{\sin C}$... Numerically $\tan 2\theta=77/36$.
Correct Answer: D