# Trigonometry questions and answers.

Extracts from this document...

Introduction

Project 3

## Trigonometry

### By David Timan

For Mr. Orr’s grade 11 U Math Class

January 6, 2003

- Find the length of the missing side in WXY to one decimal place.
Cosine law: c2 = a2 + b2 – 2ab cos C c = a2 + b2 – 2ab cos C c = 212 + 242 – 2(21)24 cos 40° c = 441 + 576 – 1008 (0.766044) c = 441 + 576 – 772.173 c = 244.827 c = 15.647 The missing side is 15.6 cm long. |

- Find F to one decimal place.
Cosine law: c2 = a2 + b2 – 2ab cos C 282 = 322 + 302 – 2(32)30 cos C 784 = 1024 + 900 – 1920 cos C 1920 cos C = 1024 + 900 – 784 1920 cos C = 1140 cos C = 1140 / 1920 C = cos-1 .593750 C = 53.6° F is 53.6°. |

- In ABC, a = 63cm, c = 47cm, and C = 38.4°. Find A given that ABC is acute.
Sine law: (sin A) / a = (sin C) / c (sin A) / 63 = (sin 38.4) / 47 (sin A) / 63 = (.621148 / 47) sin A= (.621148 / 47) * 63 sin A= .832603 A = sin-1 .832603 A = 56.4° A is 56.4°. |

- A bridge DF is built N4°W across a river. Point E is located 75 m [west of]
^{[1]}F and DEF = 46°. What is the length of the bridge?
DF is divergent of North by 4°, hence it is also divergent of West 86°. |

Middle

Tangent ratio: tan = opp / adj

tan 50 = PR / 12

PR = 12(tan 50)

PR = 12(1.191754)

PR = 14.3 cm

## For segment RQ

Tangent ratio: tan = opp / adj

tan 38 = RQ / 12

RQ = 12(tan 38)

RQ = 12(.781285)

RQ = 9.4 cm

## For segment PQ (PQ = c)

Cosine Law: c2 = a2 + b2 – 2ab cos C

c = 9.42 + 14.32 – 2(9.4)14.3 cos 134

c = 88.36 + 204.49 – 268.84(-.694658)

c = 88.36 + 204.49 +186.75

c = 479.6

c = 21.9 cm

Line segment PQ is 21.9 cm long.

- Find the length of QR.
## For PRQ“Z” Pattern for parallel lines: SPR = PRQ PRQ = 10° ## For SQRSum of Interior Angles: A + B + C = 180° SQR = 180 – (100 + 10) SQR = 70° ## For PQRAbutting Angles: PQR = SQR + PQS PQR = 28 + 70 PQR = 98° ## For QPRSum of Interior Angles: A + B + C = 180° QPR = 180 – (98 + 10) QPR = 72° ## For Segment QRSine Law: (sin A) / a = (sin C) / c (sin 10) / 13 = (sin 72) / c c / 13 = (sin 72) / (sin 10) c = 13 * (sin 72) / (sin 10) c = 13 * 5.476916 c = 71.2 The length of line segment QR is 71.2 cm. |

- If cos(x /3) = - 3 / 2, the the possible values of x are ___ and ___.
Since the values for cosine on the unit circle are found on the x-axis, when x is equal to - 3 / 2 we can see that of the radiating arms with 30° spacing the two on either side of the 180° mark meet the circle at this x value. |

Conclusion

## For exterior angles

Sum of the exterior angles of any polygon is 360°: six equal exterior angles so 180 / 6 = 60°

## For distance between flats

Draw an imaginary line dividing off the two sides to the right of the diagram. The acute angles at the top and bottom of this triangle are found to be 30° by subtracting the exterior angle from the 90° that defines the perpendicular imaginary line created above. By sum of the interior angles of a triangle the contained angle of the imaginary triangle is found to be 120°.

Sine Law: (sin A) / a = (sin C) / c

(sin 30) / 12 = (sin 120) / c

c = (sin 120) / [(sin 30) / 12]

c = .866025 / [.5 / 12]

c = 20.8 cm

## For Distance to centre

Half of the distance between flats

The centre is 20.8 / 2 = 10.4 cm

## For height of face

Pythagorean theorem: a2 + b2 = c2

c2 = 10.42 + 242

c2 = 108.16 + 576

c2 = 684.16

c = 26.156 cm

For Apex Angle.

Let x = half of the apex angle

Tangent Ratio : tan x = 6 / 26.156

tan x = .229393

x = 13°

Since half of the apex angle is 13° the apex angle is 26°.

[1] This problem does not contain enough information to be solved so the assumption was made that “E” was due west of “F” and that “D” is the northerly point of the bridge.

This student written piece of work is one of many that can be found in our AS and A Level Modern Physics section.

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