Question

In ΔJKL, j = 27 cm, k = 82 cm and ∠L=162°. Find the area of ΔJKL, to the nearest square centimeter.

Answer 1

Answer:

$684\:\mathrm{cm^2}$

Step-by-step explanation:

The area of any triangle is equal to $A=\frac{1}{2}\cdot a\cdot b\cdot \sin C$, where $a$ and $b$ are two sides of a triangle and $C$ is the angle between them.

Plugging in given values, we have:

$A=\frac{1}{2}\cdot 27\cdot 82\cdot \sin 162^{\circ}=\boxed{684\:\mathrm{cm^2}}$