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)