Question

Question 6
If $\angle$A and $\angle$B are acute angles such that cos A = cos B, then show that $\angle$A = $\angle$B.

Answer 1

Let $\Delta$ABC in which CD $\perp$ AB.
According to the question,
cos A = cos B
$\Rightarrow$ $\frac{AD}{AC}$ = $\frac{BD}{BC}$
$\Rightarrow$ $\frac{AD}{BD}$ = $\frac{AC}{BC}$
Let $\frac{AD}{BD}$ = $\frac{AC}{BC}$ = k
$\Rightarrow AD=kBD\dots \left(i\right)$
$\Rightarrow AC=kBC\dots \left(ii\right)$
By applying pythagoras theorem in$\Delta$CAD and $\Delta$CBD; we get,
$C{D}^2=A{C}^2-A{D}^2\dots .\left(iii\right)$
and also $C{D}^2=B{C}^2-B{D}^2\dots .\left(iv\right)$
From equations (iii) and (iv) we get,
$A{C}^2-A{D}^2=B{C}^2-B{D}^2$
$\Rightarrow (kBC{)}^2-(kBD{)}^2=B{C}^2-B{D}^2$
$\Rightarrow k^2(B{C}^2-B{D}^2)=B{C}^2-B{D}^2$
$\Rightarrow k^2=1$
$\Rightarrow$ k = 1
Putting this value in equation (ii), we obtain
AC = BC
$\Rightarrow$$\angle$A = $\angle$B (Angles opposite to equal sides of a triangle are equal-isosceles triangle).