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 F_{1} = 35.07 Kg, F_{2} = 22.6 Kg,

and F_{3} = 41.72 Kg. Find the angle between F_{1} and F_{2}.

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° – ( 41.87^{0} + 28.25^{0}) = **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° – (55.3^{0} + 37.2°) = **87.5** °

c) a = 4.25 m b = 7.58 m **É****A =** 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, b^{2} = 12^{2} + 13^{2} – 2(12)(13) cos 120^{0}

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 F_{1} and F_{2}.

angle* x* = arccos 0.000067 = **89. 996 **°** = 90 ^{ }**°

4)

a) Since triangles CED and CBA are similar, we can solve for **y**:

72.5*y* = 40.3*y* + 1007.5

so 32.2*y* = 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

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