If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Let ΔABC in which CD ⊥ AB.
A/q,
cos A = cos B
⇒ AD/AC = BD/BC
⇒ AD/BD = AC/BC
Let AD/BD = AC/BC = k
⇒ AD = kBD .... (i)
⇒ AC = kBC .... (ii)
By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD2 = AC2 - AD2 …. (iii)
and also CD2 = BC2 - BD2 …. (iv)
From equations (iii) and (iv) we get,
AC2 - AD2 = BC2 - BD2
⇒ (kBC)2 - (k BD)2 = BC2 - BD2
⇒ k2 (BC2 - BD2) = BC2 - BD2
⇒ k2 = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B (Angles opposite to equal sides of a triangle are equal-isosceles triangle)
