Question

In a triangle ABC, if $a\sin A=b\sin B$, then the nature of the triangle is
A. $a > b$
B. $a < b$
C. $a = b$
D. $a + b = c$

Answer 1

Hint: To solve this question, we will use Law of sine $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. We will equate the Law of sine to some constant and then derive the value of the angle $\sin A$ and $\sin B$. Then we will substitute the values of $\sin A$ and $\sin B$ derived in the given equation $a\sin A=b\sin B$and simplify. After simplifying we will get a resultant equation which will help in determining the nature of the triangle.
Formula used:
Law of sine:
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
Complete step-by-step solution:
We are given a triangle $ABC$ such that $a\sin A=b\sin B$and we have to derive the nature of the triangle.
We will take the Law of sine and equate it to some constant $k$.
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
We will only take the first two components of the relationship $\frac{a}{\sin A}=\frac{b}{\sin B}$.
We will now determine the values of the angles.
$\begin{align}
  & \frac{a}{k}=\sin A \\
 & \frac{b}{k}=\sin B \\
\end{align}$
We will now substitute the values of angles $\sin A$ and $\sin B$in the given equation $a\sin A=b\sin B$.
$\begin{align}
  & a\times \frac{a}{k}=b\times \frac{b}{k} \\
 & {{a}^{2}}={{b}^{2}} \\
 & \sqrt{{{a}^{2}}}=\sqrt{{{b}^{2}}} \\
 & a=b
\end{align}$
The nature of the triangle$ABC$such that $a\sin A=b\sin B$, then the nature of the triangle is $a=b$. Hence the correct option is (C).
Note:
The square root of any quantity gives two values positive and negative. We choose the values accordingly. In simplifying the equation,
$\sqrt{{{a}^{2}}}=\sqrt{{{b}^{2}}}$
We will get two values of $a$ and $b$ one positive and one negative that is,
$\pm a=\pm b$
Here $a$ and $b$ are the length of the sides of the triangle. As length cannot be negative, we will choose the positive values of the sides $a$ and $b$.
The triangle will be an isosceles triangle because the two sides are equal.