Im trying to teach myself some geometry on my own, but im stuck since some weeks, with a couple of area problems of triangles. I ll really appreciate any help or advice about it.

The formulas for a triangle are:

A=1/2 bh and A=1/2 ab Sen C

The problem says:
-Find the area of a triangle that has two adyacent sides that are 5 and 4 ,and they include an angle that is 45º

I thought that the formula might be A=1/2 ab Sen C

so it would be A= 1/2 (4)(5)(sen 45) = 7.07106...

but the result on the book was = 5 square root of 2

Could I use "sen of 45" in that way? if not what is the correct procedure in order to do that problem?

There are some other similar problems that I cant understand as well:

-Find the area of a triangle that has two adjacent sides that are 8 and 12, and they include an angle that is 60º.

-Find the area of a triangle if two sides are 13 and 15, and the hight over the third side is 12.

Im stuck with those ones since some weeks, I 'll really appreaciate any help.

I thought that the formula might be A=1/2 ab Sen C

so it would be A= 1/2 (4)(5)(sen 45) = 7.07106...

but the result on the book was = 5 square root of 2

Your answer is correct, but just expressed differently. Note that sin(45) = sqrt(2)/2--verify with a table of trigonometric values. A calculator does not explicitly reveal the factor, square root of 2.

Your answer is correct, but just expressed differently. Note that sin(45) = sqrt(2)/2--verify with a table of trigonometric values. A calculator does not explicitly reveal the factor, square root of 2.

Thanks a lot, I appreaciate it.

Im trying to solve this one now:

Find the area of a triangle if two sides are 13 and 15, and the hight over the third side is 12.

Do I need to find the base in the problem? Should I use the pytagorean theorem to get the base in this problem? and if so what should be the procedure?

Dick