If $∠ A a n d ∠ B$ are acute angles such that tan A = tan B then prove that $∠ A = ∠ B$ .

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Solution

$I n △ A B C, ∠ C = 90^{o} t a n A = \frac{B C}{A C} a n d t a n B = \frac{A C}{B C} A s t a n A = t a n B \Rightarrow \frac{B C}{A C} = \frac{A C}{B C} \Rightarrow B C^{2} = A C^{2} \Rightarrow B C = A C S o, ∠ A = ∠ B$

(Angles opposite to equal sides are equal)

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If ∠A and ∠ B are acute angles such that tan A = tan B then prove that ∠A=∠B.

In △ABC, ∠C=90otan A=BCAC and tan B=ACBCAs tan A=tan B⇒BCAC=ACBC⇒BC2=AC2⇒BC=ACSo, ∠A=∠B (Angles opposite to equal sides are equal)

If $∠ A a n d ∠ B$ are acute angles such that tan A = tan B then prove that $∠ A = ∠ B$ .

$I n △ A B C, ∠ C = 90^{o} t a n A = \frac{B C}{A C} a n d t a n B = \frac{A C}{B C} A s t a n A = t a n B \Rightarrow \frac{B C}{A C} = \frac{A C}{B C} \Rightarrow B C^{2} = A C^{2} \Rightarrow B C = A C S o, ∠ A = ∠ B$

(Angles opposite to equal sides are equal)