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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?

User Theglauber
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2 Answers

4 votes

Answer:

the proof is in the picture

Explanation:

In the right triangle shown, m\angle J = 60^\circm∠J=60 ∘ m, angle, J, equals, 60, degrees-example-1
User Niraj Chauhan
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4.9k points
1 vote

Answer:


|JK|=3√(3)

Explanation:

The diagram of the triangle is drawn and attached,

From trigonometric ratios,


cos \theta =(Adjacent)/(Hypotenuse) \\cos60^\circ =(|JK|)/(|JL|)\\ 0.5=(|JK|)/(6√(3) )\\$Cross multiply$\\|JK|=0.5*6√(3)\\|JK|=3√(3)

In the right triangle shown, m\angle J = 60^\circm∠J=60 ∘ m, angle, J, equals, 60, degrees-example-1
User AsgarAli
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3.9k points