Question

In ΔJKL, the measure of ∠L=90°, JL = 24, LK = 7, and KJ = 25. What is the value of the sine of ∠K to the nearest hundredth?

Answer 1

In ΔJKL

s = JK+KL+LJ/2

= 25+7+21/2

= 28

then,

ΔJKL = √[s(s-a)(s-b)(s-c)]

= √[28(28-25)(28-24)(28-7)

= √(28×3×4×21)

= √(7056

= 84cm²

Again,

ΔJKL = 1/2×JL×LK×∠L

or, 84 = 1/2×24×7×sin x°

or, 84/168 = sin x°

or, 1/2 = sin x°

or sin 30° = sin x°

or x° = 30°

At last,

∠K+∠L+∠J = 180° (sum of angle of triangle)

or, ∠K+30°+90° = 180°

or, ∠K = 180°- 120°

or, ∠K = 60°