Posted by **Tabitha** on Thursday, January 13, 2011 at 2:02am.

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.

e.g. #1

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

#2,

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.

#5

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.

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