Question

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

Answer 1

Given :-

• ∠A and ∠B are acute angles.

• cosA = cosB

To Prove :-

• ∠A=∠B

Proof :-

Let us assume the Δ ABC in which CD⊥AB.

Given that the angles A and B are acute angles, such that,

$\leadsto\rm\red{Cos (A) = cos (B)}$

As per the angles taken, the cos ratio is written as :

$\leadsto\rm\blue{AD/AC = BD/BC}</p><p>$

Now, interchange the terms, we get

$\leadsto\rm{\pink{AD/BD = AC/BC}}$

Let take a constant value

$\leadsto\mathrm{\orange{ AD/BD = AC/BC = k}}$

Now consider the equation as

$</p><p>\leadsto\rm{\green{AD = k BD …(1)}} \\ </p><p></p><p>\leadsto\rm\orange{AC = k BC …(2)}$

By applying Pythagoras theorem in Δ CAD and Δ CBD we get,

$\leadsto\rm\pink{CD^2 = BC^2- BD^2 … (3) }\\ </p><p></p><p>\leadsto\rm\green{CD^2 =AC^2 -AD^2 ….(4)}</p><p>$

From the equations (3) and (4) we get,

$\leadsto\rm{ \red{AC^2 - AD^2= BC^2 - BD^2}}$

Now substitute the equations (1) and (2) in (3) and (4)

$\leadsto\rm{ \blue{ K^2(BC^2 - BD^2)=(BC^2 - BD^2)k^2=1}}$

Putting this value in equation, we obtain

$\leadsto\rm{\green{AC = BC}}$

Angles opposite to equal side are equal-isosceles triangle :

$\leadsto\huge{\boxed{\rm{\orange{ \bold{∠A=∠B }}}}}$

Hence Proved !!

Answer 2

Concept:

just simple answer with property of isosceles triangle!!!

Answer:

Let us consider a right ∆ABC ,∠C = 90°

now,

cos A = AC / AB

and

cos B = BC/AB

Since cos A = cos B

$\therefore \bf \: \frac{AC }{ \cancel{ AB } } = \frac{BC}{\cancel{ AB }}$

Now, in the ∆ABC, two sides AC and BC are equal!!!

•°• Their opposite angles are also equal...

•°• ∠A = ∠B