Question
The diagram shows a flagpole, $BD$, held by two ropes, $AD$ and $CD$.
$ABC$ is a straight line and angle $\angle ABD = 90^\circ$.
\(AD = 21.2 \, \text{m}, \, AB = 16.5 \, \text{m}, \, \text{and angle } \angle BCD = 48^\circ.\)
(a) Show that the height of the flagpole $BD$ is $13.3 \, \text{m}$, correct to 1 decimal place.
(b) Calculate the length of the rope $CD$.
▶️Answer/Explanation
$(\mathbf{a})~ [BD= ] \sqrt {21. 2^2- 16. 5^2}$
$13.31\ldots$
$(b)$ $17.9$ or $17.89$ to $17.91$
(a)
Pythagoras’ Theorem
$AD^2 = AB^2 + BD^2$
$(21.2)^2 = (16.5)^2 + BD^2$
$449.44 = 272.25 + BD^2$
$BD^2 = 449.44 – 272.25$
$BD^2 = 177.19$
$BD = \sqrt{177.19} \approx 13.31 \, \text{m}$
(b)
$\sin(48^\circ) = \frac{BD}{CD}$
$\sin(48^\circ) = \frac{13.3}{CD}$
$0.7431 = \frac{13.3}{CD}$
$CD = \frac{13.3}{0.7431}$
$CD \approx 17.9 \, \text{m}$
Question
The diagram shows the route of a ship that leaves a port, P.
It travels due west for 16km and then changes course to due south for 9km.
(a) Calculate the straight line distance PQ.
Answer/Explanation
Ans: 18.4
(b) Use trigonometry to calculate the bearing of P from Q.
Answer/Explanation
Ans: [0]60.4 to [0]60.73
Question
In the diagram, B is 27 metres due east of A.
C is 34 metres from A and due south of B.
(a) Using trigonometry, calculate angle ACB.
Answer/Explanation
Ans: 52.6
(b) Find the bearing of C from A.
Answer/Explanation
Ans: 127 or 127.4[…]
Question
The bearing of P from M is 107°. Work out the bearing of M from P.
▶️Answer/Explanation
If the bearing of P from M is 107°, then the bearing of M from P will be the opposite direction, which is 180° added to 107°:
Bearing of M from P = 180° + 107°
Calculating this, we have:
Bearing of M from P = 287°
Therefore, the bearing of M from P is 287°.
Question
A and B are two points marked on a map. By measuring a suitable angle, find the bearing of A from B.
▶️Answer/Explanation
$328 \pm 2$ (ie 326 to 330$)$