Question

In a triangle ABC, with the usual notations, \(\angle B = \frac{π}{3}\), \(\angle C = \frac{π}{4}\). If D divides BC internally in the ratio 1:3, then \(\frac{sin\angle BAD}{sin\angle CAD} =\)

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{\sqrt{3}}\)
• C. \(\frac{1}{\sqrt{6}}\)
• D. \(\sqrt{\frac{2}{3}}\)

Hint:

Apply the sine rule in triangles ABD and ACD and use BD : DC = 1 : 3.

Answer 1

The Correct Option is C

Step 1: Ratio theorem
For a cevian AD, $\dfrac{BD}{DC} = \dfrac{AB \sin\angle BAD}{AC \sin\angle CAD}$.

Step 2: Sides
By the sine rule $AB/AC = \sin C / \sin B = \sin 45^{\circ} / \sin 60^{\circ} = \sqrt{2}/\sqrt{3}$.

Step 3: Solve
$\dfrac{\sin\angle BAD}{\sin\angle CAD} = \dfrac13 \cdot \dfrac{AC}{AB} = \dfrac13\sqrt{\dfrac32} = \dfrac{1}{\sqrt6}$. Option (C).

Final Answer:
Option (C). \[ \boxed{\frac{1}{\sqrt{6}}} \]