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Math 5
Review worksheet 2

1.  Solve triangle PIT if P = 10, I = 107, p = 56.

DRAW A DIAGRAM OF THE TRIANGLE!!!!!
This is NOT a right triangle.
Since I have SSA, it is possible that there are 2 solutions, 1 solution, or no solution.

p/sinP = i/sinI
56/sin10 = 107/sinI
sin I = 0.332
DRAW A DIAGRAM OF CAST.
Since sine is positive, the angle could lie in quadrant 1 or 2.
sin-1(0.332) = 19° (REFERENCE ANGLE)
I = 19°  (Quadrant 1)
I = 180° - 19° = 161°  (Quadrant 2)

The question now splits.  We must consider both possible answers for I.

First, if I = 19°:

Angle T:
T = 151°, by SATT

Side t:
p/sinP = t/sinT
56/sin10 = t/sin151
t = 156

Second, if I =151:

Angle T:
T = 9°, by SATT

Side t
p/sinP = t/sinT
56/sin10 = t/sin9
t = 50

Two possible solutions :
I = 19°, T = 151°, t = 156,
I = 161°, T = 9°, t = 50.


2.  Solve triangle BGT if b = 16.8, g = 27.0, t = 37. 6

DRAW A DIAGRAM OF THE TRIANGLE.
(again, nothing suggests that it is a right triangle.)

Since it is an oblique triangle, I can use the sum of angles of a triangle theorem, the sine law, or the cosine law.
Since I have 3 sides and no angles, i will use the cosine law.
Since I have SSS, I know there is at most one solution to this question. (it is possible that there is no solution)

b2 = g2 +t2 -2gt cosB
(16.8)2 + (27.0)2 + (37.6)2  - 2(27.0)(37.6)cosB
282.24 = 729.00 + 1413.76 - 2030.40 cosB
-1860.52 = -2030.40cosB
cosB = 0.916
B = 24°
(Because cosine is positve, the angle could lie in quadrant 1 or quadrant 4.  Since we are dealing with a triangle, it is not possible to be in quadrant 4.)

I could now use the sine law, but then I would have to consider two possibilities, and only one can be correct in this case.  I will use the cosine law again, as it will give me only one result.

g2 = b2 +t2  - 2bt cosG
(27.0)2 = (16.8)2 + (37.6)2 - 2(16.8)(37.6)cosG
729.00 = 282.24 + 1413.76 - 1263.36cosG
-967.00 = -1263.36cosG
cosG = 0.765
G =40°

B + G + T = 180
(24) + (40) + T = 180
T = 116

therefore, B = 24°, G = 40°, T = 116°.


3. Solve
tan x = 3.64, -2pi ¾ x ¾ 2pi

DRAW A DIAGRAM (CAST)
Since tangent is positive, the angle could occur in either quadrant 1 or quadrant 3.
tan-1(3.64) = 1.303
Quadrant 1 : x = 1.303
Quadrant 3 : x = pi + 1.303 = 4.444

therefore, x = 1.303, 4.444
 

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