SOLUTION: Find lengths correct to the nearest integer. There is an equalaterial triangle and it cuts it down the middle to form to right triangles. The whole base of the triangle is 24 ho

Algebra ->  Triangles -> SOLUTION: Find lengths correct to the nearest integer. There is an equalaterial triangle and it cuts it down the middle to form to right triangles. The whole base of the triangle is 24 ho      Log On


   

Question 298995: Find lengths correct to the nearest integer.
There is an equalaterial triangle and it cuts it down the middle to form to right triangles. The whole base of the triangle is 24 however, if you cut it in half the base of the 2 right triangles equal 12. The bottom angle on the triangle across from the right angle equals 65 degrees. Using cosine, sin, or tangent find the values of x and y. X is the hypotuense on both right triangles and y is the leg that cuts the triangle in half. Can you please help me with this problem? Thank you
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
There is an equilaterial triangle and it cuts it down the middle to form to right triangles.
The whole base of the triangle is 24 however, if you cut it in half the base of the 2 right triangles equal 12.
The bottom angle on the triangle across from the right angle equals 65 degrees.
Using cosine, sin, or tangent find the values of x and y. X is the hypotenuse
on both right triangles and y is the leg that cuts the triangle in half.
:
You can find x using the cosine of 65,
where
x = hypotenuse
12 = side adjacent
cos(65) = 12%2Fx
.4226x = 12
x = 12%2F.4226
x = 28.39 ~ 28
and
find y using the tan of 65
where
side opposite = y
side adjacent = 12
tan(65) = y%2F12
y = 2.445*12
y = 25.73 ~ 26
:
:
check solution by finding the sine of a, it should equal 65
sin(a) = 25.73/28.39
sin(a) = .9064
a = 65.0