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Lena [83]
2 years ago
11

find the exact value given that sin A= -4/5 with A in quadrant IV, tan B=7/24 with b in quadrant III, ; cos C= -5/13 with c in q

uadrant II
Mathematics
1 answer:
Kitty [74]2 years ago
4 0

Answer:

A= 306,87°

B= 196,26°

C= 112,62

Step-by-step explanation:

sinA =-4/5

A = -53,24°

A = 306,87°

tan B = 7/24

B = 16,26°

B = 196,26°

cos C = -5/13

C = 112,62°

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if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

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------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
%  (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


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