Question

In a given $\Delta ABC$, if cos A = sin B – cos C, then the triangle is
a. equilateral triangle
b. isosceles triangle
c. right-angled triangle
d. scalene triangle

Answer 1

Hint: In order to solve this question, we will start from the given equality and then we will try to represent it in such a way that we will be able to get $\cos X+\cos Y=2\cos \left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)$ and sin 2 X = 2 sin X cos X. And then we will try to form a relationship which will give us the answer.
Complete step-by-step answer:
In this question, we have been given a triangle with the condition cos A = sin B – cos C. And we have been asked to find what type of triangle it is. To solve this question, we will first consider the given equality, that is,
cos A = sin B – cos C
Now, we know that the equality can be further written as,
cos A + cos C = sin B
Now, we know that $\cos X+\cos Y=2\cos \left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)$. So, for X = A and Y = C, we can write,
$2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=\sin B$
Now, we know that sin 2 X = 2 sin X cos X. So, for 2 X = B, we can write,
$2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\sin \dfrac{B}{2}\cos \dfrac{B}{2}.........\left( i \right)$
Now, we know that the sum of all angles of a triangle is 180˚. So, we can write, for $\Delta ABC$,
A + B + C = 180˚………(ii)
A + C = 180 – B
On dividing by 2 on both the sides, we will get,
$\dfrac{A+C}{2}=\dfrac{180-B}{2}$
And we can also write it as,
$\dfrac{A+C}{2}=90-\dfrac{B}{2}$
Therefore, by using this equality, we can write equation (i) as,
$2\cos \left( 90-\dfrac{B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\sin \dfrac{B}{2}\cos \dfrac{B}{2}$
Now, we know that $\cos \left( 90-\theta \right)\sin =\theta $. So, we can write, for $\theta =\dfrac{B}{2}$,
\[2\sin \left( \dfrac{B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\sin \dfrac{B}{2}\cos \dfrac{B}{2}\]
Now, on cancelling the like terms from both the sides, we get,
   $\cos \left( \dfrac{A-C}{2} \right)=\cos \dfrac{B}{2}$
And it can be written as below,
$\begin{align}
  & \dfrac{A-C}{2}=\dfrac{B}{2} \\
 & A-C=B \\
 & A=B+C \\
\end{align}$
Now, we will use this equality and put the values in equation (ii). So, we get,
A + A = 180˚
2A = 180 ˚
$A=\dfrac{180}{90}$
A = 90˚
Hence, we can say that, $\Delta ABC$ is a right angled triangle, as the triangle has an angle of 90˚.
So, option (c) is the correct answer.

Note: There are high possibilities that, in a hurry, we might read the given equality wrong as cos A = sin B – sin C and then we will try to apply the identity of $\sin X-\sin Y=2\cos \left( \dfrac{X+Y}{2} \right)\sin \left( \dfrac{X-Y}{2} \right)$ which will definitely give us a result but that will be wrong answer. Also, we might make mistakes while writing the given equation as cos A = sin B + cos C, which will also give us wrong answers.