Thread: Precalculus angle-side-side, please help!!

Hi, I really need help on this problem please. I realize that because it's angle side side, and one is shorter than the other, there will be two sets of solutions because there are two triangles. Please help me and show me what both areas and perimeters are! Thanks!!!

If the triangle is shown like that, then there is only one solution.

If the triangle is not shown like that, if only the two sides and the one angle are mentioned without showing a figure, then there are two solutions.

Maybe you just showed the figure of one of the two possible solutions.

For the figure that is shown,

Let's call the triangle, ABC, where
>>>angle A is 15 degrees
>>>AB = 11 ft
>>>BC = 8 ft

By Law of Sines,
8/sin(15deg) = 11/sinA
Cross multiply,
8sinA = 11sin(15deg)
sinA = 11sin(15deg) /8 = 0.355876
angle A = arcsin(0.355876) = 20.847153 degrees ----**

angle B = 180 -15 -20.847153 = 144.152847 deg

By Law of Sines again,
8/sin(15deg) = AC/sin(144.152847deg)
AC = 18.1 ft

So, perimeter of ABC = 11 +8 +18.1 = 37.1 ft -------answer.
Area = (1/2)(11)(18.1)sin(15deg) = 25.765 sq.ft -----answer.

--------------------------------
For the other triangle:

Again,
By Law of Sines,
8/sin(15deg) = 11/sinA
Cross multiply,
8sinA = 11sin(15deg)
sinA = 11sin(15deg) /8 = 0.355876 ---------positive sine value, so A is in the 1st or 2nd quadrant.

In the 1st quadrant,
angle A = arcsin(0.355876) = 20.847153 degrees ----used in the above triangle.

In the 2nd quadrant,
angle A = 180 -20.847153 = 159.152847 deg -----to be used in this triangle.

angle B = 180 -15 -159.152847 = 5.847153 deg

By Law of Sines again,
8/sin(15deg) = AC/sin(5.847153 deg)
AC = 3.15 ft

So, perimeter of ABC = 11 +8 +3.15 = 22.15 ft -------answer.
Area = (1/2)(11)(3.15)sin(15deg) = 4.484 sq.ft -----answer.