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Question: In \[\vartriangle ABC\] , if \(a\cos A = b\cos B\) , then the triangle is A) Isosceles B) Right ...

In ABC\vartriangle ABC , if acosA=bcosBa\cos A = b\cos B , then the triangle is
A) Isosceles
B) Right Angled
C) Isosceles or Right Angled
D) Right Angled Isosceles

Explanation

Solution

In such type of solution first we need to simplify the given equation and and comparing angles between them so that when we compare the angle we can get the information that which triangle is formed between the following options, it can also form different triangle at same solution given as follows

Complete step-by-step answer:
Now, in the given question we need to determine the type of triangle
In the given question it is given that
acosA=bcosBa\cos A = b\cos B
If we simplify the above as
\Rightarrow acosB=bcosA\dfrac{a}{{\cos B}} = \dfrac{b}{{\cos A}}
Now, if we need to determine the value of bb
\Rightarrow acosBcosA=b\dfrac{a}{{\cos B}}\cos A = b
Now, similarly we need to determine the value bb as in the sine function
\Rightarrow asinA=bsinB\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}
Now, we will determine the value of bb
\Rightarrow asinAsinB=b\dfrac{a}{{\sin A}}\sin B = b
Now, comparing both the value of sine and cosine function, we get
\Rightarrow acosBcosA=asinAsinB\dfrac{a}{{\cos B}}\cos A = \dfrac{a}{{\sin A}}\sin B
Simplifying the above, we get
\Rightarrow asinAcosA=asinBcosBa\sin A\cos A = a\sin B\cos B
Multiplying 22 both sides of the above equation, we get
\Rightarrow 2asinAcosA=2asinBcosB2a\sin A\cos A = 2a\sin B\cos B
The above can be simplified with the help identity and can be written as follows
Now, the above can also be written as
\Rightarrow sin2A=sin2B\sin 2A = \sin 2B
Now,
sin2A=sin2B\sin 2A = \sin 2B
\Rightarrow 2A=2B2A = 2B
Simplifying we get
\Rightarrow A=BA = B
Hence it is a Isosceles triangle
But it can also be represented as
\Rightarrow sin2A=sin2B=sin(1802B)\sin 2A = \sin 2B = \sin ({180^ \circ } - 2B)
Hence, on comparing the angles we get
\Rightarrow 2A=1802B2A = {180^ \circ } - 2B
Simplifying the above, we get
\Rightarrow A=90BA = {90^ \circ } - B
Hence, we get
\Rightarrow A+B=90A + B = {90^ \circ }
Hence, the given triangle is an Isosceles or right angled triangle

Option C is the correct answer.

Note: In the given question the triangle can be isosceles or right angle triangle because the same condition provides us different triangles at different kinds of solution as in the above if we take angle equal directly then they form isosceles triangle.