GIVE MEDAL!! In right ∆ABC, m B = 90°, m C = 40°, and BC = 10. What are the other two side lengths of the triangle? AC = 13.1, AB = 8.4 AC = 18.7, AB = 15.8 AC = 14.3, AB = 10.2 AC = 15.6, AB = 11.9

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GIVE MEDAL!! In right ∆ABC, m B = 90°, m C = 40°, and BC = 10. What are the other two side lengths of the triangle? AC = 13.1, AB = 8.4 AC = 18.7, AB = 15.8 AC = 14.3, AB = 10.2 AC = 15.6, AB = 11.9

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We can use Sine, Cosine, and Tangent to find the remaining sides..
|dw:1436967646848:dw|
\(\sf Sine=\dfrac{Opposite}{Hypotenuse}\) \(\sf Cosine=\dfrac{Adjacent}{Hypotenuse}\) \(\sf Tangent = \dfrac{Opposite}{Adjacent}\) Here, we are given the adjacent side, and we want to find the opposite and the hypotenuse.

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Hehe, it would be if you got it yourself :P
o..k
umm..
|dw:1436968060414:dw|
So to find the hypotenuse, we can use cosine. \(\sf Cosine=\dfrac{Adjacent}{Hypotenuse}\) Plug in what we know: \(\sf cos(90^o) = \dfrac{10}{x}\)
Hmm..I think that's right..I'm not sure about which angle to use..I haven't done this in a while..
its ok
Okay, 90 degrees is wrong..
Hold on a minute
k
Okay, I think we use 40 degrees..
So: \(\sf cos(40^o) = \dfrac{10}{x}\) Type in \(\sf cos(40^o)\) into your calculator, tell me what you get.
i dont have a scientific one ..its ok ill guess but thx for help ill give tht medal :)
You can use google, just type in "cos 40 degrees"
kk
a big decimal so ill put 0.766
Yep, so we have: \(\sf 0.766 = \dfrac{10}{x}\) Solve for 'x'
so i times the decimal and 10?
or divide
Well, you can switch 'x' and 0.766 around. \(\sf x = \dfrac{10}{0.766}\) So just divide
13.1?
it was a big decimal so i round it and got tht
Sorry, my internet just went out for a few seconds there.
Yep, 13.1 is corret.
correct*
That's our hypotenuse.
its a!!!
Yes, I think that's correct.
thx for the help :)
Np

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