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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?

Respuesta :

Answer:

[tex]|JK|=3\sqrt{3}[/tex]

Step-by-step explanation:

The diagram of the triangle is drawn and attached,

From trigonometric ratios,

[tex]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}[/tex]

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copperdog1011

Answer:

the proof is in the picture

Step-by-step explanation:

Ver imagen copperdog1011

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