Question

In a triangle ABC, $\angle A={30}^{0}andb:c=(\sqrt{3}-1):1then:(\angle C-\angle B)$ has the value equal to

• A. ${60}^0$
• B. ${30}^0$
• C. ${22.5}^0$
• D. ${45}^0$

Answer 1

The correct option is A ${60}^0$

Given:-

$\frac{b}{c}=\sqrt{3}-1$ and $\angle A={30}^0$

Since $A+B+C={180}^0$

Hence $B+C={150}^0$.................[1]

Using $sin$ rule:

$\frac{c}{\sin C}=\frac{b}{\sin B}$

$⟹\frac{b}{c}=\frac{\sin B}{\sin C}$........[2]

Now

$\frac{b}{c}=\sqrt{3}-1=\sqrt{3}-1\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2}{\sqrt{3}+1}=\frac{2/2\sqrt{2}}{(\sqrt{3}+1)/2\sqrt{2}}=\frac{1/\sqrt{2}}{(\sqrt{3}+1)/2\sqrt{2}}$

From $\left[2\right]$:

$\frac{\sin B}{\sin C}=\frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\frac{\sin {45}^0}{\sin {105}^0}$

Since these angle values satisfy equation $1$.

Hence:

$\angle C-\angle B={105}^0-{45}^0={60}^0$