For problems 1 and 2, determine how many solutions there are for each triangle. You do not have to solve the triangle.
1. A = 29°, a = 13, c = 27
2. A = 100.1°, a = 20, b = 11
For problems 3-6, solve each triangle using the Law of Sines. If there is no solution, write “no solution.” Round each answer to the nearest tenth.
3. A = 41°, B =61°, c = 19
4. A = 125.4°, a = 33, b = 41
5. A = 76°, a = 15, b = 5
6. A = 97.5°, a = 13, b = 9
Math - Reiny, Thursday, January 13, 2011 at 8:29am
Your exercise probably deals with the
"ambiguous case", that is, the information usually consists of two sides and a non-contained angle.
make sketches marking the given information, even though in reality the triangle may not be possible.
by the sine law:
Sin C/27 = sin 29°/13
sinC = 1.0069
this is not possible since the sine of any angle must be between -1 and +1.
So, no solution is possible
sinB/11 = sin100.1/20
sinB = .54147
angle B = 32.784
but the sine is + in quadrants I or II
so angle B could also have been 180-32.784 = 147.726°.
However, a triangle cannot have two obtuse angles, so the second case is not possible.
Find the third side and angle in the usual way.
sinB/5 = sin76/15
sinB = .3234
angle B = 18.9° or 161.1°
In this case the sum of 161.1 + 76 is already > 180, so we have to go with the solution of
angle B = 18.9°.
The rest of the question is routine.