Question

If in triangle $ABC$, $\cos A=\frac{\sin B}{2\sin C}$ then the triangle is
A. Equilateral.
B. Isosceles.
C. Right angled.
D. None of these.

Answer 1

Hint: To solve this question, we will write the given equation $\cos A=\frac{\sin B}{2\sin C}$ in terms of $\sin B$ and consider it as first equation. We will use angle sum property of triangle and write the equation in terms of the angle $B$. Then we will substitute the value of angle $B$derived in the first equation and simplify it using trigonometric formulas. After this we will get the resultant equation which will be property of some triangle which in terms will help in determining the type of triangle.
Formula used:
\[\begin{align}
  & \sin (A+B)=\sin A\cos B+\cos A\sin B \\
 & \sin (A-B)=\sin A\cos B-\cos A\sin B \\
\end{align}\]
Complete step-by-step solution:
We are given a triangle $ABC$such that $\cos A=\frac{\sin B}{2\sin C}$and we have to determine the type of the triangle $ABC$ is.
We will use the given equation $\cos A=\frac{\sin B}{2\sin C}$ and write it in terms of $\sin B$,
$2\sin C\cos A=\sin B\,\,\,\,......(i)$
Now we will use angle sum property of the triangle,
$\begin{align}
  & A+B+C=\pi \\
 & B=\pi -(A+C).....(ii)
\end{align}$
We will now substitute the equation (ii) in equation (i),
$\begin{align}
  & 2\sin C\cos A=\sin (\pi -(A+C)) \\
 & 2\sin C\cos A=\sin (A+C)
\end{align}$
We will now use the formula of \[\sin (A+B)\],
\[\begin{align}
  & 2\cos A\sin C=\sin A\cos C+\cos A\sin C \\
 & 2\cos A\sin C-\cos A\sin C-\sin A\cos C=0 \\
 & \cos A\sin C-\sin A\cos C=0
\end{align}\]
We know that \[\cos A\sin C-\sin A\cos C\] is the formula of $\sin (C-A)$, so
\[\begin{align}
  & \sin (C-A)=0 \\
 & \sin (C-A)=\sin 0 \\
 & C-A=0 \\
 & C=A
\end{align}\]
Two angles of the triangle are equal. We will now draw the triangle,

We know the isosceles property of the triangle according to which sides opposite to equal angles are equal which means that if angles $C=A$ then their opposite will be also equal that is $AC=AB$.
And if two sides are equal then that angle is an isosceles triangle.
The triangle $ABC$ such that$\cos A=\frac{\sin B}{2\sin C}$ is an isosceles triangle. Hence the correct option is (B).
Note:
 We could have also solved this question using cosine rule and Law of sine.
Using Law of sine $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, we will derive the value of $\frac{\sin B}{\sin C}$.
\[\begin{align}
  & \frac{b}{\sin B}=\frac{c}{\sin C} \\
 & \frac{b}{c}=\frac{\sin B}{\sin C} \\
\end{align}\]
We will now use cosine rule $\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$ and equation we derived \[\frac{b}{c}=\frac{\sin B}{\sin C}\] in $\cos A=\frac{\sin B}{2\sin C}$.
$\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\frac{b}{2c}$
Solving this equation we will get equation $c=a$. As both sides are equal then the triangle will be isosceles.