If $∠ A$ and $∠ B$ are acute angles such that tanA = tanB, the prove that $∠ A = ∠ B$ .

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Solution

In $∆ ABC, ∠ C = 90 °$ ,

$tanA = \frac{BC}{AC} and tanB = \frac{AC}{BC}$

$As, tanA = tanB$
$\Rightarrow \frac{BC}{AC} = \frac{AC}{BC} \Rightarrow {BC}^{2} = {AC}^{2} \Rightarrow BC = AC So, ∠ A = ∠ B (Angles opposite to equal sides are equal)$

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If ∠A and ∠B are acute angles such that tanA = tanB, the prove that ∠A=∠B.

In ∆ABC, ∠C=90°, tanA=BCAC andtanB=ACBC As, tanA=tanB ⇒BCAC=ACBC⇒BC2=AC2⇒BC=ACSo, ∠A=∠B Angles opposite to equal sides are equal

If $∠ A$ and $∠ B$ are acute angles such that tanA = tanB, the prove that $∠ A = ∠ B$ .

In $∆ ABC, ∠ C = 90 °$ ,

$tanA = \frac{BC}{AC} and tanB = \frac{AC}{BC}$

$As, tanA = tanB$
$\Rightarrow \frac{BC}{AC} = \frac{AC}{BC} \Rightarrow {BC}^{2} = {AC}^{2} \Rightarrow BC = AC So, ∠ A = ∠ B (Angles opposite to equal sides are equal)$