Question

In \( \triangle ABC \), \( \angle B = 60^\circ \) and \( \angle A = 75^\circ \). If a point \( D \) divides \( BC \) in the ratio \( 2:3 \), then \( \sin \angle BAD : \sin \angle CAD = \)

• A. \( \sqrt{2} : \sqrt{3} \)
• B. \( \sqrt{3} : 2 \)
• C. \( \sqrt{3} : \sqrt{2} \)
• D. \( 3 : \sqrt{2} \)

Hint:

Apply the sine rule to triangles \( ABD \) and \( ACD \). The ratio of sines of the angles \( \angle BAD \) and \( \angle CAD \) will be proportional to the ratio of the segments \( BD \) and \( DC \) and the sines of the angles \( \angle B \) and \( \angle C \).

Answer 1

The Correct Option is A

In \( \triangle ABC \), \( \angle A = 75^\circ \) and \( \angle B = 60^\circ \).
The sum of angles in a triangle is \( 180^\circ \), so \( \angle C = 180^\circ - (75^\circ + 60^\circ) = 180^\circ - 135^\circ = 45^\circ \).
Let \( BD : DC = 2 : 3 \).
Let \( BD = 2k \) and \( DC = 3k \) for some \( k>0 \).
Using the sine rule in \( \triangle ABD \): \( \frac{BD}{\sin \angle BAD} = \frac{AD}{\sin \angle B} \implies \sin \angle BAD = \frac{BD \sin B}{AD} = \frac{2k \sin 60^\circ}{AD} = \frac{k \sqrt{3}}{AD} \).
Using the sine rule in \( \triangle ACD \): \( \frac{DC}{\sin \angle CAD} = \frac{AD}{\sin \angle C} \implies \sin \angle CAD = \frac{DC \sin C}{AD} = \frac{3k \sin 45^\circ}{AD} = \frac{3k}{AD \sqrt{2}} \).
The ratio \( \sin \angle BAD : \sin \angle CAD = \frac{k \sqrt{3}}{AD} : \frac{3k}{AD \sqrt{2}} = \frac{\sqrt{3}}{1} : \frac{3}{\sqrt{2}} = \sqrt{3} \sqrt{2} : 3 = \sqrt{6} : 3 = \sqrt{2} \sqrt{3} : \sqrt{3} \sqrt{3} = \sqrt{2} : \sqrt{3} \).