Question

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Answer 1

SOLUTION
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)

Answer 2

We are given that,

$\qquad \rm {• cos \: A = cos \: B }$

________

We have to show,

$\qquad \rm {• ∠A = ∠B }$

________

Construction,

$\dag$ Draw a triangle ABC where C = 90°.

__________

Solution

We know,

$\qquad \rm {• cos \: A = cos \: B }$

$\implies { \rm \dfrac{B}{H} = \dfrac{B}{H} }$

$\implies { \rm \dfrac{AC}{AB} = \dfrac{BC}{AB} }$

$\implies rm{ AC \times \cancel {AB} = BC \times \cancel {AB} }$

$\implies { \rm AC = BC }$

Now , we know that if two sides of triangle are equal than it is isosceles triangle and we also know that in isosceles triangle, if two opposite sides are equal than their angle will also be equal.

$\implies { \rm \angle A= \angle B }$

Thus, prooved!