Question

If in a triangle ABC, $a\cos A=b\cos B$, then the triangle is
A. Isosceles triangle
B. Right-angled triangle
C. Isosceles or right-angled triangle
D. Right-angled isosceles

Answer 1

Hint: In order to solve this question, we should know the concept of sine law of triangle, which states that for triangle ABC, having angles as A< B and C and sides opposite to these angles as a, b, and c respectively, we can say $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=K$. We can find the answer to this question by using this concept.

Complete step-by-step answer:
In this question, we have been asked to find the type of triangle which satisfies $a\cos A=b\cos B$. To solve this question, we should know about the concept of sine law of a triangle ABC, having angles as A, B and C and sides opposite to these angles as a, b and c respectively, we can say, $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=K$.

So, we can write the equality $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=K$ as,
$\dfrac{\sin A}{a}=K$ and $\dfrac{\sin B}{b}=K$
And we can further write it as,
$a=\dfrac{\sin A}{K}$ and $b=\dfrac{\sin B}{K}$
Now, we will put the values of A and B in the given equality, that is, $a\cos A=b\cos B$. So, we get,
$\dfrac{\sin A}{K}\cos A=\dfrac{\sin B}{K}\cos B$
Now, we know that the common term on both sides of the equation can be cancelled out. So, we get,
$\sin A\cos A=\sin B\cos B$
Now, we know that any number or fraction can be multiplied to both sides of the equality. So, on multiplying both sides of the equality by 2, we get,
$2\sin A\cos A=2\sin B\cos B$
Now, we know that $2\sin \theta \cos \theta$ can be expressed as $\sin 2\theta$. So, we can say for $\theta =A,2\sin A\cos A=\sin 2A$ and for $\theta =B,2\sin B\cos B=\sin 2B$. Therefore, we get the above equality as,
$\begin{aligned} & \sin 2A=\sin 2B \\\ & \Rightarrow \sin 2A-\sin 2B=0 \\\ \end{aligned}$
Now, we know that $\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)$. So, for x = 2A and y = 2B, we get, $\dfrac{x-y}{2}=\dfrac{2A-2B}{2}=A-B$ and $\dfrac{x+y}{2}=\dfrac{2A+2B}{2}=A+B$. Therefore, we get the equality as,
$2\cos \left( A+B \right)\sin \left( A-B \right)=0$
Now, we know that for the above equality to satisfy either $\cos \left( A+B \right)=0$ or $\sin \left( A-B \right)=0$. We know that $\cos \theta =0$ for $\theta =90{}^\circ$ and $\sin \theta =0$ for $\theta =0{}^\circ$. So, we can write,
$\cos \left( A+B \right)=\cos 90{}^\circ$ or $\sin \left( A-B \right)=\sin 0{}^\circ$
And it can be expressed further as,
$A+B=90{}^\circ$ or $A-B=0{}^\circ$
$\Rightarrow A+B=90{}^\circ$ or $A=B$
Now, we know that for a right-angled triangle, the sum of 2 angles = third angle = $90{}^\circ$. So, we can say $A+B=90{}^\circ$ is a condition for a right-angled triangle, right angled at C.
We also know that if 2 angles of a triangle are equal, then the pair of sides opposite to these angles will also be equal, that means if A = B, then a = b. Therefore, triangle ABC is an isosceles triangle.
Hence, from both the above conclusions, we can say that triangle ABC is either a right-angled triangle or an isosceles triangle. Therefore, option C is the correct answer.

Note: While solving this question, the possible mistakes we can make is while choosing the correct option. We might choose option D, which is a single case of both the possible type of triangles. Also, we need to remember that $2\sin \theta \cos \theta =\sin 2\theta$ and $\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)$.