Question

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

Answer 1

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that

cos A = cos B

⇒ `("AD")/("AC") = ("BD")/("BC")`            ...(1)

We have to prove ∠A = ∠B.

To prove this, let us extend AC to P such that BC = CP.

From equation (1), we obtain

`("AB")/("BD") = ("AC")/("BC")`

⇒ `("AD")/("BD") = ("AC")/("CP")`           ...(By construction, we have BC = CP)          ...(2)

By using the converse of B.P.T,

CD || BP

⇒ ∠ACD = ∠CPB         ...(Corresponding angles)             ...(3)

And, ∠BCD = ∠CBP          ...(Alternate interior angles)               …(4)

By construction, we have BC = CP

∴ ∠CBP = ∠CPB              ...(Angle opposite to equal sides of a triangle)       …(5)

From equations (3), (4) and (5), we obtain

∠ACD = ∠BCD               …(6)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD               ...[Using equation (6)]

∠CDA = ∠CDB               ...[Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

Alternatively,

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that,

cos A = cos B

⇒ `("AD")/("AC") = ("BC")/("BC")`

⇒ `("AD")/("BD") = ("AC")/("BC")`

Let `("AD")/("BD") = ("AC")/("BC") = k`

⇒ AD = k × BD                    …(1)

And, AC = k × BC               …(2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD² = AC² − AD²           …(3)

And, CD² = BC² − BD²               …(4)

From equations (3) and (4), we obtain

AC² − AD² = BC² − BD²

⇒ (k BC)² − (k BD)² = BC² − BD²

⇒ k² (BC² − BD²) = BC² − BD²

⇒ k² = 1

⇒ k = 1

Putting this value in equation (2), we obtain

AC = BC

⇒ ∠A = ∠B               ...(Angles opposite to equal sides of a triangle)

Answer 2

∠A and ∠B are acute angles

Cos A = cos B S.T ∠A = ∠B

Let us consider right angled triangle ACB.

We have cos A = `"adjacent side"/"Hypotenuse"`

= `("AC")/("AB")`

cos B = `("BC")/("AB")`

cos A = cos B

`("AC")/("AB") = ("BC")/("AB")`

AC = BC

∠A = ∠B