Solving: (According to the picture)(Using the trigonometric functions)
[tex]Cos\:60^0 = \frac{Opposite\:leg}{hypotenuse} [/tex]
[tex] \frac{1}{2} = \frac{K}{24} [/tex]
multiply cross
[tex]2*K = 24*1[/tex]
[tex]2K = 24[/tex]
[tex]K = \frac{24}{2} [/tex]
[tex]\boxed{K = 12}\end{array}}\qquad\quad\checkmark[/tex]
Substituting the side found in the image, we form another triangle, with the unknown "J" to be found.
We find two ways of finding the value of the "J" side.
First Way:
Solve for the Pythagorean Theorem: "In a triangle rectangle the sum of the squares of the legs is equal to the square of the hypotenuse"
[tex]J^2 + 12^2 = 24^2[/tex]
[tex]J^2 + 144 = 576[/tex]
[tex]J^2 = 576 - 144[/tex]
[tex]J^2 = 432[/tex]
[tex]J = \sqrt{432} [/tex]
Make the Least Common Multiple (432)
[tex]432\:|2[/tex]
[tex]216\:|2[/tex]
[tex]108\:|2[/tex]
[tex]54\:|2[/tex]
[tex]27\:|3[/tex]
[tex]9\:|3[/tex]
[tex]3\:|3[/tex]
[tex]1\:|\underline{2^2*2^2*3^2*3}[/tex]
Replace in square root
[tex]J = \sqrt{432} [/tex]
[tex]J = \sqrt{2^2*2^2*3^2*3} [/tex]
[tex]J = 2*2*3\sqrt{3} [/tex]
[tex]\boxed{J = 12 \sqrt{3} }\end{array}}\qquad\quad\checkmark[/tex]
Second Way:(According to the picture)(Using the trigonometric functions)
[tex]Cos\:30^0 = \frac{Opposite\:leg}{hypotenuse} [/tex]
[tex] \frac{ \sqrt{3} }{2} = \frac{J}{24} [/tex]
multiply cross
[tex]2*J = 24* \sqrt{3} [/tex]
[tex]2J = 24 \sqrt{3} [/tex]
[tex]J = \frac{24 \sqrt{3} }{2} [/tex]
[tex]\boxed{J = 12 \sqrt{3} }\end{array}}\qquad\quad\checkmark[/tex]
Answer:
[tex]\boxed{J = 12 \sqrt{3} }\end{array}}\qquad\quad\checkmark[/tex]
[tex]\boxed{K = 12}\end{array}}\qquad\quad\checkmark[/tex]
