Question

The angle of elevation of a tower from a point
L is 62°. From a point K, 50 m further from the tower, the
angle of elevation is 47°. (Let the height of the tower be h.)
a Use the sine rule in ΔKTL to show that:
TL =
b Use trigonometry in ΔLMT to show that:
TL =
c Hence, show that h =
d Calculate the height, h, of the tower, correct to one decimal place.
15 From the top of a cliff, the angles of depression of two boats
at sea 0.5 km apart are 55° and 33°. (Let the height of the cliff
be h.)
a Show that the height of the cliff is:
h =
b Hence, calculate the height, correct to the nearest metre.

Answer 1

Answer on Question #44256, Math, Trigonometry

The angle of elevation of a tower from a point $L$ is $62{}^\circ$ . From a point $K$ , $50 \, \text{m}$ further from the tower, the angle of elevation is $47{}^\circ$ . (Let the height of the tower be $h$ .)

a Use the sine rule in $\Delta$ KTL to show that: TL =

b Use trigonometry in $\Delta$ LMT to show that: TL =

c Hence, show that $h =$

d Calculate the height, h, of the tower, correct to one decimal place.

15 From the top of a cliff, the angles of depression of two boats at sea $0.5\mathrm{km}$ apart are $55{}^{\circ}$ and $33{}^{\circ}$ . (Let the height of the cliff be h.)

a Show that the height of the cliff is: h =

b Hence, calculate the height, correct to the nearest metre.

Solution:

Angles of elevation and depression $(\Phi)$ are formed by the horizontal lines that a viewer's lines of sight form to an object.

For the first problem we have drawing:

TM=h (height of the tower), MK=50 m (distance from a point K to tower)

a) In trigonometry, sine rule is an equation relating the lengths of the sides to the sines of its angles. According to the law

$$\frac {\sin A}{a} = \frac {\sin B}{b} = \frac {\sin C}{c}$$

where $a, b,$ and $c$ are the lengths of the sides of a triangle, and $A, B,$ and $C$ are the opposite angles.

Hence in $\Delta KTL$ from sine rule: $\frac{\sin\angle KTL}{LK} = \frac{\sin\angle TKL}{TL}$

$\angle KTL = 180{}^{\circ} - \angle TKL - (180{}^{\circ} - \angle TLM) = 180{}^{\circ} - 47{}^{\circ} - (180{}^{\circ} - 62{}^{\circ}) = 15{}^{\circ}$ . From here we obtain

TL=LK\*sin 47° sin 15°

b) in $\Delta LMT\angle TML = 90{}^{\circ}$ , hence $TL = \frac{TM}{\sin\angle TLM} = \frac{TM}{\sin 62{}^{\circ}} = \frac{h}{\sin 62{}^{\circ}}$

c) from b) we obtain $h = TL * \sin 62{}^\circ$ , also from $\Delta KMT \frac{TM}{MK} = \tan 47{}^\circ$ ,

hence $h = MK^{*} \tan 47{}^{\circ}$ .

d) $h = MK^{*} \tan 47{}^{\circ} = 50^{*}1.072 = 53.6$ (m)

Second problem:

A-first boat, B-second boat, AB=0.5 km distance between boats

$\angle OCB = 33{}^{\circ}$ , $\angle OCA = 55{}^{\circ}$ the angles of depression of two boats.

Lines CO and DA are parallel, hence $\angle BAC = 180{}^{\circ} - \angle OCA = 180{}^{\circ} - 55{}^{\circ} = 125{}^{\circ}$ .

In $\Delta ABC$ from sine rule: $\frac{\sin\angle BAC}{BC} = \frac{\sin\angle ACB}{AB}$ , and $BC = \frac{\sin\angle BAC}{\sin\angle ACB} * AB$ .

$\angle ACB = \angle OCA - \angle OCB = 22{}^{\circ}$ . In $\Delta CDB \angle CDB = 90{}^{\circ}$ , $\angle DBC = 90{}^{\circ} - 33{}^{\circ} = 57{}^{\circ}$ .

Hence

a) $h = CD = BC^*$ $\cos \angle DBC = \frac{\sin \angle BAC}{\sin \angle ACB} * AB * \cos \angle DBC$

b) from a) obtain $h = 0.5 * \frac{\sin 125{}^\circ}{\sin 22{}^\circ} * \cos 57{}^\circ = 0.595$ (km)

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