Question
Question: In \[\vartriangle ABC\] , if \(a\cos A = b\cos B\) , then the triangle is A) Isosceles B) Right ...
In △ABC , if acosA=bcosB , then the triangle is
A) Isosceles
B) Right Angled
C) Isosceles or Right Angled
D) Right Angled Isosceles
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=bcosB
If we simplify the above as
⇒ cosBa=cosAb
Now, if we need to determine the value of b
⇒ cosBacosA=b
Now, similarly we need to determine the value b as in the sine function
⇒ sinAa=sinBb
Now, we will determine the value of b
⇒ sinAasinB=b
Now, comparing both the value of sine and cosine function, we get
⇒ cosBacosA=sinAasinB
Simplifying the above, we get
⇒ asinAcosA=asinBcosB
Multiplying 2 both sides of the above equation, we get
⇒ 2asinAcosA=2asinBcosB
The above can be simplified with the help identity and can be written as follows
Now, the above can also be written as
⇒ sin2A=sin2B
Now,
sin2A=sin2B
⇒ 2A=2B
Simplifying we get
⇒ A=B
Hence it is a Isosceles triangle
But it can also be represented as
⇒ sin2A=sin2B=sin(180∘−2B)
Hence, on comparing the angles we get
⇒ 2A=180∘−2B
Simplifying the above, we get
⇒ A=90∘−B
Hence, we get
⇒ A+B=90∘
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.