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5.01

1. solve the triangle.
a = 33°, a = 19, b = 14 (1 point)

b = 23.7°, c = 143.3°, c ≈23.3
b = 23.7°, c = 123.3°, c ≈17.5
cannot be solved
b = 23.7°, c = 123.3°, c ≈29.2
2. solve the triangle.
b = 73°, b = 15, c = 10 (1 point)

c = 39.6°, a = 67.4°, a ≈14.5
cannot be solved
c = 44.8°, a = 62.4°, a ≈14.5
c = 39.6°, a = 67.4°, a ≈20.3
3. state whether the given measurements determine zero, one, or two triangles.
c = 30°, a = 28, c = 14 (2 points)

one
zero
two
4. two triangles can be formed with the given information. use the law of sines to solve the triangles.
c = 67°, a = 21, c = 20 (2 points)

a = 14.9°, b = 98.1°, b = 21.5; a = 165.1°, b = 81.9°, b = 21.5
a = 75.1°, b = 37.9°, b = 13.3; a = 104.9°, b = 8.1°, b = 3.1
a = 75.1°, b = 37.9°, b = 30; a = 104.9°, b = 8.1°, b = 30
a = 14.9°, b = 98.1°, b = 18.6; a = 165.1°, b = 81.9°, b = 18.6
5. the given measurements may or may not determine a triangle. if not, then state that no triangle is formed. if a triangle is formed, then use the law of sines to solve the triangle, if it is possible, or state that the law of sines cannot be used.
b = 126°, c = 7, b = 12 (2 points)

c = 25.8°, a = 28.2°, a ≈6.5
c = 28.2°, a = 25.8°, a ≈6.5
no triangle is formed.
the triangle cannot be solved with the law of sines.
6. two weather tracking stations are on the equator 146 miles apart. a weather balloon is located on a bearing of n 35°e from the western station and on a bearing of n 23°e from the eastern station. how far is the balloon from the western station? (2 points)
an obtuse triangle is shown with base from point west to point east of 146 miles. line segments extend from both the east and west points to the third point of the triangle entitled balloon. the angle opposite the hypotenuse shows ninety degrees in addition to a north east bearing of 23 degrees and the west angle shows a bearing to the triangle of 35 degrees north east.