Question

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

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Answer 1

Let us assume the triangle ABC in which CD⊥AB

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

Cos (A) = cos (B)

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

AD/AC = BD/BC

Now, interchange the terms, we get

AD/AC = BD/BC

Let take a constant value

AD/AC = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = K BC …(2)

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

CD2=BC2-BD2…(3)

CD2=AC2-AD2..(4)

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

AC2-AD2=BC2-BD2

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

k2(BC-BD2)=(BC2-BD2)K2=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

Hope it's helpful

Answer 2

Answer:

Let us assume the triangle ABC in which CD⊥AB

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

Cos (A) = cos (B)

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

AD/AC = BD/BC

Now, interchange the terms, we get

AD/AC = BD/BC

Let take a constant value

AD/AC = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = K BC …(2)

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

CD2=BC2-BD2…(3)

CD2=AC2-AD2..(4)

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

AC2-AD2=BC2-BD2

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

k2(BC-BD2)=(BC2-BD2)K2=1

Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

Hope this helps you