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Maths-

General

Easy

Question

In a right triangle, the tangent of one of the two acute angles is $fraction numerator square root of 3 over denominator 3 end fraction$ . What is the tangent of the other acute angle?

$\frac{- \sqrt{3}}{3}$
$\frac{- 3}{\sqrt{3}}$
$\frac{\sqrt{3}}{3}$
$\frac{3}{\sqrt{3}}$

The correct answer is: $fraction numerator 3 over denominator square root of 3 end fraction$

Hint:
Tangent of an angle = $fraction numerator O p p o s i t e space s i d e over denominator space A d j a c e n t space s i d e space end fraction$
Step-by-step solution:-
In the adjacent diagram, let ΔABC be the right angle triangle with ∠BAC = 90° and ∠ ABC as the given acute angle with Tangent $fraction numerator square root of 3 over denominator 3 end fraction$ .
Now, we know that-
Tangent of an angle = $fraction numerator O p p o s i t e space s i d e over denominator space A d j a c e n t space s i d e space end fraction$
∴ Tangent of ∠BAC = $fraction numerator A C over denominator A B end fraction$
∴ $fraction numerator square root of 3 over denominator 3 end fraction$ = $fraction numerator A C over denominator A B end fraction$
∴ AC = √3 & AB = 3
Now, Tangent of other acute angle i.e. ∠BAC-
= $fraction numerator O p p o s i t e space s i d e over denominator space A d j a c e n t space s i d e space end fraction$

= $fraction numerator A C over denominator A B end fraction$
= $fraction numerator 3 over denominator square root of 3 end fraction$
Final Answer:
Option D i.e. $fraction numerator 3 over denominator square root of 3 end fraction$ is the correct Option.

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