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: Understanding the Concept
Angle A is \(180^{\circ} - 60^{\circ} - 45^{\circ} = 75^{\circ}\). The point D divides BC in the ratio \(BD : DC = 1 : 3\).

Step 2: Sine rule in two triangles
In \(\triangle ABD\): \(\dfrac{BD}{\sin\angle BAD} = \dfrac{AD}{\sin B}\).
In \(\triangle ACD\): \(\dfrac{DC}{\sin\angle CAD} = \dfrac{AD}{\sin C}\).
Divide the first by the second:
\[ \frac{BD \sin\angle CAD}{DC \sin\angle BAD} = \frac{\sin C}{\sin B} \Rightarrow \frac{\sin\angle BAD}{\sin\angle CAD} = \frac{BD}{DC}\cdot\frac{\sin B}{\sin C} \]

Step 3: Substitute
\[ = \frac13 \times \frac{\sin 60^{\circ}}{\sin 45^{\circ}} = \frac13 \times \frac{\sqrt3/2}{1/\sqrt2} = \frac13\sqrt{\frac32} = \frac{1}{\sqrt6} \]
Option (B), \(1/\sqrt3\), would result from leaving out the \(\sin 45^{\circ}\) term.

Final Answer:
The ratio is \(\frac{1}{\sqrt6}\), option (C). \[ \boxed{\frac{1}{\sqrt{6}}} \]