Question

\(sin(3sin^{-1}(\frac{1}{5})) =\)

• A. \(\frac{74}{125}\)
• B. \(\frac{71}{125}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{1}{2}\)

Hint:

Apply the sine rule in two sub-triangles and use the ratio BD to DC.

Answer 1

The Correct Option is B

Step 1: Use areas:
Triangles ABD and ADC have the same height from A, so $\frac{\text{Area}(ABD)}{\text{Area}(ADC)} = \frac{BD}{DC} = \frac13$.

Step 2: Write areas with sines:
Area(ABD) = $\frac12 AB\cdot AD\sin\angle BAD$ and Area(ADC) = $\frac12 AC\cdot AD\sin\angle CAD$. So $\frac{AB\sin\angle BAD}{AC\sin\angle CAD} = \frac13$.
With $\frac{AB}{AC} = \sqrt{\frac23}$, we get $\frac{\sin\angle BAD}{\sin\angle CAD} = \frac13\sqrt{\frac32} = \frac{1}{\sqrt6}$. Option (C).

Final Answer:
$\frac{1}{\sqrt6}$. \[ \boxed{\frac{1}{\sqrt{6}}} \]