- #1
H.M. Murdock
- 34
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Im trying to teach myself some geometry on my own, but I am 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 can't 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.
Thanks a lot in advance.
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 can't 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.
Thanks a lot in advance.