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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(all content of the MathRoom Lessons © Tammy the Tutor; 2004 - ).