Question

The angles $A, B$ and $C$ of a triangle $ABC$ are in A.P. If $b : c =\sqrt {3} : \sqrt {2}$ then the angle $A$ is

• A. 30$^\circ$
• B. 15$^\circ$
• C. 75$^\circ$
• D. 45$^\circ$

Answer 1

Answer: (C)

75$^\circ$

Answer 2

The correct answer is C:$75\degree$
Given data:
In $\triangle{ABC}$
$b\ratio{c}=\sqrt{3}\ratio\sqrt{2}$
we know that,
$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$-(i)
the angles A,B,C are in A.P(Given)
$\therefore 2B=A+C$
we know that, $\angle{A}+\angle{B}+\angle{C}=180\degree$
⇒$2\angle{B}+\angle{B}=180\degree$
⇒$\angle{B}=60\degree$
From equation (i), $\frac{b}{c}=\frac{sinB}{sin{C}}$
⇒$\frac{\sqrt{3}}{\sqrt{2}}=\frac{sin60\degree}{sinc}$
⇒$\frac{\sqrt{3}}{\sqrt{2}}=\frac{\frac{\sqrt{3}}{2}}{sinc}$
⇒$sinc=\frac{1}{\sqrt{2}}$
⇒$\angle{C}=45\degree$
$\therefore \angle{A}+\angle{B}+\angle{C}=180\degree$
⇒$\angle{A}=180\degree-60\degree-45\degree$
⇒$A=75\degree$