Question

In the right triangle shown, m\angle J = 60^\circm∠J=60 ∘ m, angle, J, equals, 60, degrees and JL = 6\sqrt{3}JL=6 3 ​ J, L, equals, 6, square root of, 3, end square root. How long is JKJKJ, K?

Answer 1

Answer:

$|JK|=3\sqrt{3}$

Step-by-step explanation:

The diagram of the triangle is drawn and attached,

From trigonometric ratios,

$cos \theta =\frac{Adjacent}{Hypotenuse} \\cos60^\circ =\frac{|JK|}{|JL|}\\ 0.5=\frac{|JK|}{6\sqrt{3} }\\ Cross multiply \\|JK|=0.5*6\sqrt{3}\\|JK|=3\sqrt{3}$