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

2 Answers

7 votes

Answer:

342

Explanation:

see image

In ΔJKL, j = 27 cm, k = 82 cm and ∠L=162°. Find the area of ΔJKL, to the nearest square-example-1
User Mony
by
4.7k points
3 votes

Answer:


684\:\mathrm{cm^2}

Explanation:

The area of any triangle is equal to
A=(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=(1)/(2)\cdot 27\cdot 82\cdot \sin 162^(\circ)=\boxed{684\:\mathrm{cm^2}}

User Flovilmart
by
3.7k points