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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 Cc =    a2 + b2 – 2ab cos Cc =    212 + 242 – 2(21)24 cos 40°c =    441 + 576 – 1008 (0.766044)c =    441 + 576 – 772.173c =    244.827c = 15.647 The missing side is 15.6 cm long. Find F to one decimal place.  Cosine law: c2 = a2 + b2 – 2ab cos C282 = 322 + 302 – 2(32)30 cos C784 = 1024 + 900 – 1920 cos C1920 cos C = 1024 + 900 – 7841920 cos C = 1140cos C = 1140 / 1920C = cos-1 .593750C = 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) * 63sin A= .832603A = sin-1 .832603A = 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

For segment PR

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.

1. Find the length of QR.

## For PRQ

“Z” Pattern for parallel lines: SPR = PRQ

PRQ = 10°

## For SQR

Sum of Interior Angles: A + B + C = 180°

SQR = 180 – (100 + 10)

SQR = 70°

## For PQR

Abutting Angles: PQR = SQR + PQS

PQR = 28 + 70

PQR = 98°

## For QPR

Sum of Interior Angles: A + B + C = 180°

QPR = 180 – (98 + 10)

QPR = 72°

## For Segment QR

Sine 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.

1. 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

[is] 12 cm and the verticle hieght is 24 cm.  Determine the measure of the apex angle of each face.

## 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.

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