Question

The area of a triangle is 12 sq. cm. Two sides are 6 cm and 12 cm. The included angle is

• A. $co{s}^{-1}\left(\frac{1}{3}\right)$
• B. $co{s}^{-1}\left(\frac{1}{6}\right)$
• C. $si{n}^{-1}\left(\frac{1}{6}\right)$
• D. $si{n}^{-1}\left(\frac{1}{3}\right)$

Answer 1

Answer: (D)

$si{n}^{-1}\left(\frac{1}{3}\right)$

Consider the given triangle

Let $a=6cm$ and $b=12cm$

Therefore the included angle is $\theta =\angle C$

$area=\frac{\left(height\right)\left(base\right)}{2}$

$area=\frac{bh}{2}$

now $h=asin\theta$

therefore

$area=\frac{basin\theta }{2}$

$12=\frac{\left(12\right)\left(6\right)sin\theta }{2}$

$sinC=sin\theta =\frac{1}{3}$

Hence $C=\theta =si{n}^{-}{}^1\left(\frac{1}{3}\right)$

The answer is D