CAL-PREP TRIG REVIEW-1 |
F/ Trigonometry
Reminder:
Notation Info: The inverse trig functions (sin 1, cos 1, and tan 1,) are called
arcsin, arccos and arctan in these lessons to avoid confusing them with the reciprocal trig functions which are cosecant, secant and cotangent.
1/ Right Triangles
Example:
For right triangles use the definitions of the trig functions (Soh Cah Toa) to find sides.
A = 90° 47° = 43°
b = 12 sin 47° = 8.78
a = 12 cos 47° = 8.18
Example: If given only sides -- to find angles use arcsin, arccos or arctan
depending on which ratio is given.
so angle B = 60.24°
angle A = 90° 60.24° = 29.76°
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2/ Oblique Triangles (no right angle!!)
2a/ Law of Sines:
use when given either two angles & one side; or two sides & one angle opposite a given side.
finding sides:
, use the appropriate pair.
finding an angle:
then use arcsin.
If the angle is obtuse, find the acute angle then subtract from 180.
Example: We're given 2 angles and 1 side. Solve the triangle:
angle C = 180° ( 30° + 45° ) = 105°
by the law of sines, ,
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2b/ Law of Cosines:
use if given two sides & contained angle, or given all three sides, no angles.
finding sides:
a² = b² + c² 2bc cos A | b² = a² + c² 2ac cos B | c² = a² + b² 2ab cos C |
then take square root.
finding an angle:
cos A = | cos B = | cos C = |
then use arccos to find the angle measure.
Find the biggest angle first -- then use law of sines to find the next angle.
Use subtraction to find the 3rd angle.
Example: Solve the triangle ABC if a = 2 cm, b = 3 cm, c = 4 cm.
We always solve for the largest angle first, since it might be obtuse.
Solving for angle C with
so angle C = arccos ( 0.25) = 104.48°
We find the second unknown angle with the Law of Sines.
sin B = 0.726176
therefore angle B = arcsin 0.726176 = 46.57°
and angle A = 180° (104.48° + 46.57° ) = 28.95°
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Make diagrams for all questions.
1) Use the Law of Sines to solve the triangles ABC:
a) b = 15.43, a = 21.76, B = 28.25°
b) B = 37.20, b = 16.4 cm a = 22.3 cm
c) a = 4.25 m b = 7.58 m A = 22.67°
2) Use the Law of Cosines to solve the triangles ABC:
a) a = 3, b = 5, c = 7 | b) a = 12, c = 13, angle B = 120° |
3) A force triangle consists of three forces F1 = 35.07 Kg, F2 = 22.6 Kg,
and F3 = 41.72 Kg. Find the angle between F1 and F2.
4) The shadows of two vertical poles measure 72.5 cm and 40.3 cm.
If one pole is 25 cm taller than the other:
a) Find the angle of elevation of the sun. | b) Find the length of each pole. |
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Solutions
1) a) b = 15.43, a = 21.76, angle B = 28.25°
angle A = arcsin 0.667494 so angle A = 41.87°
angle C = 180° ( 41.870 + 28.250) = 109.88°
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b) angle B = 37.2 °, b = 16.4 cm a = 22.3 cm
angle A = arcsin 0.822107, so angle A = 55. 3 °
angle C = 180° (55.30 + 37.2°) = 87.5 °
c) a = 4.25 m b = 7.58 m ÉA = 22.67 °
angle B = arcsin 0.687413 = 43.43 °
angle C = 180 ° (43.43 ° + 22.67 °) = 113.9 °
c = 10.1 m.
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2) a) a = 3, b = 5, c = 7
cos C =
angle C = arccos ( 0.5) = 120°
sin B = 0.61859 so angle B = 38. 21°
angle A = 180 ° (120 ° + 38.21 °) = 21.79 °
b) a = 12, c = 13, ÉB = 120 °
b 2 = a 2 + c 2 2ac cos B so, b2 = 122 + 132 2(12)(13) cos 1200
sin C = 0.519775
angle C = arcsin 0.519775 = 31.32 °
angle A = 180 ° (120 ° + 31.32 °) = 28. 68 °
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3) Find the angle between F1 and F2.
angle x = arccos 0.000067 = 89. 996 ° = 90 °
4)
a) Since triangles CED and CBA are similar, we can solve for y:
72.5y = 40.3y + 1007.5
so 32.2y = 1007.5, and y = 31.29
We're looking for either angle D or angle A since they're both the angle of elevation of the sun.
tan D = = 0.776427 so angle D = arctan 0.776427 = 37. 83 °
b) We've already found the shorter pole when we found y, so, the shorter pole is 31. 29 cm tall and the longer pole is 31.29 + 25 = 56. 29 cm.
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