Question

In a triangle ABC , with usual notations if \(\mathrm{a}=4, \mathrm{~b}=8, \angle \mathrm{C}=60^{\circ}\), then the value of \(\angle \mathrm{B}\) and the ratio \(\cos \mathrm{A}: \cos \mathrm{C}\) respectively are,

• A. \(\frac{\pi}{4},1:\sqrt{3}\)
• B. \(\frac{\pi}{2},\sqrt{3}:1\)
• C. \(\frac{\pi}{2},2:\sqrt{3}\)
• D. \(\frac{\pi}{6},\sqrt{3}:2\)

Hint:

Find side c with the cosine rule, then use the sine rule to find the angles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Given \(a=4,\ b=8,\ C=60^{\circ}\). Find \(c\) using \(c^{2}=a^{2}+b^{2}-2ab\cos C\).

Step 2: Find c
\(c^{2}=16+64-2(4)(8)\dfrac{1}{2}=48\), so \(c=4\sqrt{3}\).

Step 3: Find the Angles
\(\sin A=\dfrac{a\sin C}{c}=\dfrac{4\cdot\frac{\sqrt{3}}{2}}{4\sqrt{3}}=\dfrac{1}{2}\), so \(A=30^{\circ}\) and \(B=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}=\dfrac{\pi}{2}\).

Step 4: Ratio
\(\cos A:\cos C=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}:1\), option (B).

Final Answer:
Angle B is \(\dfrac{\pi}{2}\) and the ratio \(\cos A:\cos C\) is \(\sqrt{3}:1\). \[ \boxed{\text{(B) }\dfrac{\pi}{2},\ \sqrt{3}:1} \]