What’s Sin A cos A and tan A for a triangle with the opposite of 15 and a hypotenuse of 39?

3 Answers
Feb 23, 2018

SinA=5/13
CosA=12/3
TanA=5/12

Explanation:

First you have to know some formulas. First, I will show how to find a leg of a triangle.
a^2+b^2=c^2
Then turn this around and get b^2=c^2-a^2
c^2 is the hypotenuse and a^2 is the opposite.
So now you plug in the numbers.
b^2=39^2-15^2
And after simplifying, b^2=1296
And b=√1296=36
Now SinA is opposite over hypotenuse. So SinA=15/39=5/13
CosA is adjacent over hypotenuse. So CosA=36/39=12/13
TanA is opposite over adjacent. So TanA=15/36=5/12

Feb 23, 2018

"see explanation"

Explanation:

"find the third side (adjacent) using "color(blue)"Pythagoras' theorem"

rArr"adjacent "=sqrt(39^2-15^2)=36

rArrsinA="opposite"/"hypotenuse"=15/39=5/13

cosA="adjacent"/"hypotenuse"=36/39=12/13

tanA="opposite"/"adjacent"=15/36=5/12

Feb 23, 2018

sin hatA = a / b = 5/13

cos hatA = b / c = 12/13

tan hatA = a / b = 5 / 12

Explanation:

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Given : a = 15, c = 39

As per Pythagoras theorem,

c^2 = a^2 + b^2

:. b^2 = c^2 - a^2 ==39^2 - @15^2 = 1296

b = sqrt1296 = 36

sin hatA = a / b = 15/39 = 5/13

cos hatA = b / c = 36/39 = 12/13

tan hatA = a / b = 15 / 36 = 5 / 12