Thursday
March 23, 2017

Post a New Question

Posted by 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 - , 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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question