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alexgriva [62]
4 years ago
5

What are the values of x and y x=9/4 y=3/4x=9/4 y=15/4x=15/4 y=5/4x=3/4 y=15/4

Mathematics
1 answer:
laiz [17]4 years ago
6 0
The complete question in the attached figure

we know that

in the triangle BDC
cos alfa=3/y----------> equation 1
sin alfa=x/y----------> equation 2


in the triangle ABD
cos alfa=4/5--------> equation 3
sin alfa=3/5---------> equation 4

then

Equal equation 1 and 3
3/y=4/5---------> y/3=5/4----------> y=15/4

Equal equation 2 and 4
x/y=3/5-------> x=3*y/5------------> 3*15/(4*5)-----> x=9/4

the answer is
<span>x=9/4 y=15/4</span>

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~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill

\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill

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