MAT1275 College Algebra and Trigonometry

Spring 2018 | Professor Kate Poirier

Page 2 of 16

Test #3 Review

Solve the triangle △ABC.

A = ?

C = 32°

B = 81°

a = ?

b = 13

c = ?

32° + 81° + A = 180°

113° + A = 180

A = 67° ✓

13 / sin (81) = c / sin (32)

13 sin (32) / sin (81) = c

6.888950435 / .9876883406 = c

6.9748 = c ✓

a / sin (67) = 13 / sin (81)

a = 13 sin (67) / sin (81)

a = 12.1157 ✓

A = 67°

C = 32°

B = 81°

a = 12.1157

b = 13

c = 6.9748 ✓

Test #3 Solutions

Solve the triangle △ABC.

A = 90°

B = 25°

C = ?°

a = 8

b = ?

c = ?

90° + 25° + C = 180°

115° + C = 180°

C = 65° ✓

a / sin (A) = b / sin (B)

8 / sin (90) = b / sin (25)

8 * sin (25) / sin (90) = b

3.380946094 / 1 = b

3.3809 = b ✓

b / sin (B) = c / sin (C)

3.3809 / sin (25) = c / sin (65)

3.3809 * sin (65) / sin (25) = c

7.2505 = c ✓

A = 90°

B = 25°

C = 65°

a = 8

b = 3.3809

c = 7.2505 ✓

Extra Credit. Anna. Z.

 

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