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2 years ago

5

Can someone solve this for me?

1 answer:

Helga [31]2 years ago

5 0

well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so

$\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare$

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3 years ago

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kifflom [539]

Given:

In triangle KLM, KL = 123 cm and measure of angle K is 35 degrees.

To find:

The length of the side KM to the nearest tenth of a centimeter.

Solution:

In a right angle triangle,

$\cos \theta =\dfrac{Base}{Hypotenuse}$

In the given right triangle KLM,

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