Two ships leave port P at the same time. One ship sails 60 k m on a bearing of 030 ° to position A . The other ship sails 100 k m on a bearing of 110 ° to position B . Both ships took the same time, t hours, to reach their P positions. The speed of the faster ship was 20 k m / h . Write down the value of t .

According to the question, Two ships leave port P at the same time. One ship sails 60 k m on a bearing of 030 ° to position A . The other ship sails 100 k m on a bearing of 110 ° to position B . Both ships took the same time, t hours, to reach their P positions. The speed of the faster ship was 20 k m / h . We know that P B is longer than A P . Since both the ships took the same time, it means that the ship travelling to B was the faster ship as it had to cover a larger distance. So, Time = Distance Speed Time = 100 20 = 5 h o u r s Therefore, the value of t is 5 hours .

Two ships leave port P at the same time. One ship sails 60 k m on a bearing of 030 ° to position A . The other ship sails 100 k m on a bearing of 110 ° to position B . Both ships took the same time, t hours, to reach their P positions. The speed of the faster ship was 20 k m / h . Write down the speed of the slower ship.

According to the question, Two ships leave port P at the same time. One ship sails 60 k m on a bearing of 030 ° to position A . The other ship sails 100 k m on a bearing of 110 ° to position B . Both ships took the same time, t hours, to reach their P positions. The speed of the faster ship was 20 k m / h . We know that P B is longer than A P . Since both the ships took the same time, it means that the ship travelling to B was the faster ship as it had to cover a larger distance. So, Time = Distance Speed Time = 100 20 = 5 h o u r s As given that the time taken by both the ships is same then, Time taken by slower ship is 5 h o u r s Now, Speed = Distance Time Speed of slower ship = 60 5 = 12 k m / h

The diagram shows a ladder of length 8 m leaning against a vertical wall. Use trigonometry to calculate h . Give your answer correct to 2 significant figures.

We have, The diagram shows a ladder of length 8 m leaning against a vertical wall. We know that the sine of an acute angle in a right-angled triangle is the ratio between the side opposite the angle and the hypotenuse of the triangle. Then, sin 56 ° = h 8 0 . 83 = h 8 h = 0 . 83 × 8 = 6 . 6 m Therefore, the value of h = 6 . 6 m

The diagram shows a ladder of length 8 m leaning against a vertical wall. Calculate the length of A B .

We have, The diagram shows a ladder of length 8 m leaning against a vertical wall. We know that the cosine of an angle is the length of the adjacent side divided by the length of the hypotenuse. Then, cos 28 ° = A B 8 0 . 88 = A B 8 A B = 0 . 88 × 8 = 7 . 04 c m Therefore, the length of A B = 7 . 04 c m .

In triangle A B C , A B = 6 c m , B C = 4 c m and angle B C A = 65 ° . Calculate angle C A B .

We have, In triangle A B C , A B = 6 c m , B C = 4 c m and angle B C A = 65 ° . Apply sine rule in ∆ A B C A B sin 65 ° = B C sin C A B 6 sin 65 ° = 4 sin C A B Cross multiply the given equation. Take the value of sin 65 up to 3 decimal points. Now, 6 × sin C A B = 4 × sin 65 ° sin C A B = 4 × 0 . 906 6 sin C A B = 3 . 624 6 sin C A B = 0 . 604 ∠ C A B = sin - 1 0 . 604 ∠ C A B = 37 . 2 ° Therefore, the angle C A B = 37 . 2 °

In triangle A B C , A B = 6 c m , B C = 4 c m and angle B C A = 65 ° . Calculate the area of the triangle A B C .

We have, In triangle A B C , A B = 6 c m , B C = 4 c m and angle B C A = 65 ° . Apply sine rule in ∆ A B C A B sin 65 ° = B C sin C A B 6 sin 65 ° = 4 sin C A B Cross multiply the given equation. Take the value of sin 65 up to 3 decimal points. Now, 6 × sin C A B = 4 × sin 65 ° sin C A B = 4 × 0 . 906 6 sin C A B = 3 . 624 6 sin C A B = 0 . 604 ∠ C A B = sin - 1 0 . 604 ∠ C A B = 37 . 2 ° Now, in ∆ A B C 65 ° + 37 . 2 ° + ∠ A B C = 180 ° (Angle sum property) ∠ A B C = 77 . 8 ° Then, the area of the triangle A B C = 1 2 × B C × B A × sin 77 . 8 ° = 1 2 × 4 × 6 × 0 . 977 = 11 . 7 c m 2

Karen Morrison and Nick Hamshaw Solutions for Exercise 14: Examination practice

Author:Karen Morrison & Nick Hamshaw

Karen Morrison Mathematics Solutions for Exercise - Karen Morrison and Nick Hamshaw Solutions for Exercise 14: Examination practice

Attempt the practice questions from Exercise 14: Examination practice with hints and solutions to strengthen your understanding. Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition solutions are prepared by Experienced Embibe Experts.

Questions from Karen Morrison and Nick Hamshaw Solutions for Exercise 14: Examination practice with Hints & Solutions

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 15 - Scale Drawings, Bearings and Trigonometry>Examination practice>Q 9

Two ships leave port $P$ at the same time. One ship sails $60 k m$ on a bearing of $030 °$ to position $A$ . The other ship sails $100 k m$ on a bearing of $110 °$ to position $B$ .

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Calculate the bearing of $B$ from $A$ .

EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 15 - Scale Drawings, Bearings and Trigonometry>Examination practice>Q 9

Two ships leave port $P$ at the same time. One ship sails $60 k m$ on a bearing of $030 °$ to position $A$ . The other ship sails $100 k m$ on a bearing of $110 °$ to position $B$ .

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Both ships took the same time, $t$ hours, to reach their $P$ positions. The speed of the faster ship was $20 k m / h$ . Write down the value of $t$ .

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 15 - Scale Drawings, Bearings and Trigonometry>Examination practice>Q 9

Two ships leave port $P$ at the same time. One ship sails $60 k m$ on a bearing of $030 °$ to position $A$ . The other ship sails $100 k m$ on a bearing of $110 °$ to position $∠ C A B = 180 ° - 66.6 ° = 113.4 °$ .

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Both ships took the same time, $t$ hours, to reach their $P$ positions. The speed of the faster ship was $20 k m / h$ . Write down the speed of the slower ship.

EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 15 - Scale Drawings, Bearings and Trigonometry>Examination practice>Q 1

The diagram shows a ladder of length $8 m$ leaning against a vertical wall.

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