Question

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

Answer 1

`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`

Answer 2

`"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`

Answer 3

`sin hatA = a / b = 5/13`

`cos hatA = b / c = 12/13`

`tan hatA = a / b = 5 / 12`

Explanation:

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`