Question

If \[\angle A\] and \[\angle B\] are acute angles such that \[\cos A = \cos B\] then prove that \[\angle A = \angle B\]

Answer 1

Hint: In this question, first of all construct a right-angled triangle with \[\angle A\] and \[\angle B\] are acute. Then equate the values of \[\cos A\] and \[\cos B\] to get a relation between the sides of the triangle and to prove it as a right-angled isosceles triangle. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Let us consider a right angle \[\Delta ABC\] where \[\angle C = {90^0}\], \[\angle A\] and \[\angle B\] are acute angles as shown in the figure:

Now, consider \[\cos A\] in \[\Delta ABC\]
\[\cos A = \dfrac{{{\text{adjacent side of }}\angle {\text{A}}}}{{{\text{Hypotenuse}}}} = \dfrac{{AC}}{{AB}}.............................\left( 1 \right)\]
Also, consider \[\cos B\] in \[\Delta ABC\]
\[\cos B = \dfrac{{{\text{adjacent side of }}\angle {\text{B}}}}{{{\text{Hypotenuse}}}} = \dfrac{{BC}}{{AB}}.............................\left( 2 \right)\]
Given \[\cos A = \cos B\]. So, from equations (1) and (2) we have
\[
   \Rightarrow \dfrac{{AC}}{{AB}} = \dfrac{{BC}}{{AB}} \\
  \therefore AC = BC \\
\]
We know that in a triangle, if one angle is right angle and the two sides other the hypotenuse are equal then the triangle is said to be right angled isosceles triangle.
As in right angled isosceles triangle the acute angles are equal, we have
\[\angle A = \angle B\]
Hence proved.

Note:In a right-angled isosceles triangle two angles are equal, the two sides corresponding to these angles are equal and one of the angles is the right angle. The cosine of an angle is the length of the adjacent side divided by the length of the hypotenuse.