Trigonometry help

yOu can do it!!gudLucK!!

I know what you mean am in 11th grade and I failed 2yrs of math I sux Well I dont know what to tell tou other than keep paying more attention in school because its only going to get harder…(wait thats no help sorry)…lol…

I already learned about sin, cot, and tan. I am only in the 7th grade

soh cah toa

sin = opposite over adjacent cot = adjacent over hypotenuse tan = opposite over adjacent…

here’s a website that may help

http://www.hyper-ad.com/tutoring/math/trig/trig_review.htm

feel free to funmail me if you need more help

and calm down, it’s not that hard…

I hope this renders right…

A |
| \ \ | \
|…\ …
B C D

Ok suppose you have a triangle that looks like triangle ACD (an accute triangle), and suppose you know the length of the line segments AC, AD, and CD. Imagine a right triangle ABD that encompasses it.

AB^2 + BD^2 = AD^2 (AD is known)

…also, AB^2 + AC^2 = BC^2 (AC is known)

…also BC = BD - CD (CD is known)

So, we have 3 equations, and 3 unknowns. The unknowns are AB, BC, and BD. Solve the system of equations, and you will have AB, which is the height you are looking for.

Cot=Co-Tangent I think you need because= Co-sine and you dont have to use any of those bimjob is right use the Pythagorean theorem a^2+b^2=c^2
where “a” is one side “b” is the other side and “c” is the hypotonuse. Which ever one you need to know (a and b are interchangeable the only one that needs to be specific is the hypotonuse (c)) just solve for that letter if you need anymore “individualized” help go ahead and send me a funmail ill be able to help you more spefifically there!

do you still need help?