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liq [111]
3 years ago
15

HELP PLEASE! I don't understand and have been stuck on this problem for a whole hour. Please explain. Also, none of the answers

may be correct.

Mathematics
2 answers:
dybincka [34]3 years ago
8 0
Traingle GIH and FGH are similar triangles because they both have angle H and a right angle. Since the angles in a triangle must add to 180 you know that angle their third angle is also the same. You can caculate GH through Pythagorean theorem to get around 21.9. Now you that you can use similar triangle ratios such as GF/GH is equal to IG/IH. Now plug in the numbers to get GF/21.9=15/16. GF is 20.5. Then using Pythagorean theorem FI is sqrt(20.5^2-15^2) or 13.9. 
Lyrx [107]3 years ago
5 0
The altitude is the mean proportional between the left and right parts of the hyptenuse

\cfrac{FI}{GI} =  \cfrac{GI}{HI} \ \ \Rightarrow \ \  \cfrac{FI}{15} =  \cfrac{15}{16} \ \ \Rightarrow \ \ FI= \cfrac{15 \cdot 15 }{16} \approx  14.1

Answer: 14.1
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Question in picture<br><br>A) 16/63<br><br>B)-16/63<br><br>C) 63/16<br><br>D) -63/16
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\tan(x + y) =  -  \frac{63}{16}


EXPLANATION


We were given that,

\csc(x)  =  \frac{5}{3}

This implies that,

\sin(x)  =  \frac{3}{5}

We use the Pythagorean identity

\sin^{2} (x)  +  \cos^{2} (x)= 1
to get,


\cos(x)  =  \sqrt{1 - ( { \frac{3}{5} })^{2}}  =  \frac{4}{5}


We were also given that,


\cos(y)  =  \frac{5}{13}

This means that,


\sin(y)  =  \sqrt{1 -  {( \frac{5}{13}) }^{2} }  =  \frac{12}{13}

This is because,


0 <  \: x \:  <  \frac{\pi}{2}


0 <  \: y \:  <  \frac{\pi}{2}

This angles are in the first quadrant so we pick the positive values.

\tan(x + y)  =  \frac{ \sin(x + y) }{ \cos(x + y) }


\tan(x + y)  =  \frac{ \sin(x ) \cos(y)   +  \sin(y)  \cos(x) }{ \cos(x) \cos(y)  -  \sin(x)  \sin(y) }



\tan(x + y)  =  \frac{  \frac{3}{5}   \times  \frac{5}{13}  +   \frac{12}{13}   \times  \frac{4}{5}  }{  \frac{4}{5}  \times  \frac{5}{13}   -   \frac{3}{5}  \times  \frac{12}{13}  }



\tan(x + y) =  -  \frac{63}{16}

The correct answer is D
4 0
3 years ago
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