Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
In a $\triangle ABC$, $\angle B=\frac{\pi}{3}$ and $\angle C=\frac{\pi}{4}$, also $D$ divides $BC$ internally in the ratio $1:3$, then $\frac{\sin\angle BAD}{\sin\angle CAD}$ is equal to:
$\frac{1}{\sqrt{6}}$
$\frac{1}{3}$
$\frac{1}{\sqrt{3}}$
None of these
Step-by-Step Solution
Key Concept: Apply sine rule in multiple triangles and combine ratios to find relationships between unknown angles.
From $\frac{AD}{BD} = \frac{\sin 60°}{\sin\theta}$ and $\frac{AD}{DC} = \frac{\sin 45°}{\sin\phi}$, we form the ratio $\frac{DC}{BD} = \frac{AD}{BD} \cdot \frac{DC}{AD} = \frac{\sin 60°}{\sin 45°} \cdot \frac{\sin\phi}{\sin\theta}$. Substituting values: $\frac{3}{1} = \frac{\sqrt{3}/2}{1/\sqrt{2}} \cdot \frac{\sin\phi}{\sin\theta}$, which yields $\frac{\sin\theta}{\sin\phi} = \frac{1}{\sqrt{6}}$.
Correct Answer: 1