Question

In ΔJKL, l = 380 cm, j = 620 cm and ∠K=30°. Find the area of ΔJKL, to the nearest square centimeter.

Answer 1

58,900cm²

Explanation:

Area of the triangle = 1/2absin thets

Area of the triangle = 1/2* i * j sin 30

Area of the triangle = 1/2 * 380 * 620 sin 30

Area of the triangle = 380 * 310 * 0.5

Area of the triangle = 380*155

Area of the triangle = 58,900cm²

hence the area of the triangle is 58,900cm²