Question 6:

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

Answer:

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

AD/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 = 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)

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