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
$⇒\frac{AD}{AC}=\frac{BD}{BC}.....(1)$

We have to prove $∠A = ∠B$.

To prove this, let us extend AC to P such that BC = CP.
From equation (1), we obtain
$\frac{AD}{BD}=\frac{AC}{BC}$

$⇒\frac{AD}{BD}=\frac{AC}{CP}......(2)$
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.
$∴∠CAD = ∠CBD$
$⇒ ∠A = ∠B$

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Alternatively,

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

It is given that,
cos A = cos B
$⇒\frac{AD}{AC}=\frac{BD}{BC}$

$⇒ \frac{AD}{BD}=\frac{AC}{BC}$

Let$\frac{AD}{BD}=\frac{AC}{BC}=k$
$⇒ AD = k BD … (1)$
And,$AC = k BC … (2)$
Using Pythagoras theorem for triangles CAD and CBD, we obtain
$(\text{CD}) ^2 =(\text{ AC})^ 2 -(\text{ AD})^ 2 … (3)$

And, $(\text{CD}) ^2 =(\text{ BC}) ^2 - (\text{BD})^ 2 … (4)$

From equations (3) and (4), we obtain
$(AC)^ 2 -( AD)^ 2 = (BC)^ 2 - (BD )^2$
$⇒ (k\ BC) ^2 - (k\ BD)^ 2 = (BC)^ 2 - (BD)^ 2$
$⇒ k^ 2 (BC ^2 - BD ^2 ) = BC ^2 - BD^ 2$
$⇒ k ^2 = 1$
$⇒ k = 1$
Putting this value in equation (2), we obtain
AC = BC
$⇒ ∠A = ∠B$ (Angles opposite to equal sides of a triangle)