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11.5 Applications of Right-And Non-Right-Angled Trigonometry

This document contains worked solutions to problems involving applications of right- and non-right-angled trigonometry. It includes finding the area of a triangle, calculating lengths and bearings using trigonometry, and finding angles between faces of a tetrahedron. The solutions demonstrate using trigonometric functions like cosine, sine and arccos to solve problems.

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Aanya Ralhan
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382 views2 pages

11.5 Applications of Right-And Non-Right-Angled Trigonometry

This document contains worked solutions to problems involving applications of right- and non-right-angled trigonometry. It includes finding the area of a triangle, calculating lengths and bearings using trigonometry, and finding angles between faces of a tetrahedron. The solutions demonstrate using trigonometric functions like cosine, sine and arccos to solve problems.

Uploaded by

Aanya Ralhan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Additional exercise

11.5 Applications of right- and


non-right-angled trigonometry
1 Find the area of a triangle with sides 7 cm, 8 cm and 9 cm.

a Find the length AC.

b Find the length AB.

c Find the area of triangle ABD.

3 B is 60 km from A on a bearing of 086°. C is 80 km from B on a bearing of 147°. Find the


distance and bearing of C from A.

4 This regular tetrahedron has edges of length 8 cm.

Calculate the angle between two faces.

5 In triangle ABC, AB = 7 cm, AC = 8 cm and BC = 10 cm.

a Without using a calculator, find the exact value of the cosine of angle BAC.

b Without any further calculation, explain how you know that this angle is acute.

© Oxford University Press 2019 Additional exercise 1


Additional exercise

Answers

1 Angle between the sides of length 7 cm and 8 cm is given by

 72 + 82 − 92 
arccos  =  73.3...°
 2×7×8 
1
So area = × 7 × 8 × sin73.3...° ≈ 26.8 cm2
2

AC 7.3
2 a = AC 5.303... km ≈ 5.30 km
⇒=
sin36° sin126°

b AB
= 5.303...2 + 6.12 − 2 × 5.303... × 6.1 ×
= cos 54° 5.225... km ≈ 5.23 km

CD 7.3
c = CD 2.788... km
⇒=
sin18° sin126°

1
So area of ABD = × 7.3 × (6.1 + 2.788...) × sin36°
2
≈ 19.1 km2

3 Distance = 602 + 802 − 2 × 60 × 80 cos119°

= 121.0... km ≈ 121 km
sin ∠BAC sin119°
∠BAC is given by =
80 121.0...
⇒ ∠BAC = 35.3...°
So bearing ≈ 86 + 35 = 121°

4 The perpendicular height of each triangular=


face 8 sin 60° . So the angle between each face is
 ( 8 sin60° )2 + ( 8 sin60° )2 − 82 
given by arccos   ≈ 70.5° .
 2 × 8 sin60° × 8 sin60° 
 

72 + 82 − 102 13
5 a cos=
∠BAC =
2×7×8 112

b Because the value of the cosine is positive, the angle must be acute.

© Oxford University Press 2019 2

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