5 Marks Question

Question 15 Marks

As observed from the top of a lighthouse, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation. $\big[\text{Use}\sqrt{3}=1.732\big]$

Answer

Let AB be the light house and let C and D be the positions of the ship.
Let AD = x, CD = y

In $\triangle\text{BDA},$
$\frac{\text{x}}{100}=\cot60^\circ$
$\text{x}=\frac{100}{\sqrt{3}}\text{m}$
Similarlly in $\triangle\text{BCA},\frac{\text{x}+\text{y}}{100}=\cot30^\circ$
$\Rightarrow(\text{x}+\text{y})=100\sqrt{3}\text{m}$
$\text{y}=(\text{x}+\text{y})-\text{x}$
$=\Big(100\sqrt{3}-\frac{100}{\sqrt{3}}\Big)\text{m}=\Big(\frac{200}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\Big)\text{m}$
$=115.46\text{m}$
The distance travelled by the ship during the period of observation = 115.46m.

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Question 25 Marks

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40m vertically above X, the angle of elevation is 45°. Find the height of tower. $\big[\text{Take}\sqrt{3}=1.732\big]$

Answer

Given that PQ is a vertical tower.
Let h be the height of the tower.
In right $\triangle\text{QPX},$
$\tan60^\circ-\frac{\text{PQ}}{\text{XP}}$
$\Rightarrow\sqrt{3}=\frac{\text{h}}{\text{XP}}$
$\Rightarrow\text{XP}=\frac{\text{h}}{\sqrt{3}}\text{m}\dots(\text{i})$
In right $\triangle\text{QRY},$
$\tan45^\circ=\frac{\text{QR}}{\text{YR}}$
$\Rightarrow1=\frac{(\text{h}-40)}{\text{YR}}$
$\Rightarrow\text{YR}=(\text{h}-40)\text{m}\dots(\text{ii})$
Since XP = YR, From (i) and (ii), we have
$\frac{\text{h}}{\sqrt{3}}=\text{h}-40$
$\Rightarrow\text{h}=\sqrt{3}\text{h}-40\sqrt{3}$
$\Rightarrow\sqrt{3}\text{h}-\text{h}=40\sqrt{3}$
$\Rightarrow\text{h}=\frac{40\sqrt{3}}{(\sqrt{3}-1)}$
On rationalising we get,
$\text{h}=\frac{40\sqrt{3}}{(\sqrt{3}-1)}\times\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$\Rightarrow\text{h}=\frac{120+40\sqrt{3}}{2}$
$=40\bigg(\frac{3+\sqrt{3}}{2}\bigg)$
$=20(3+\sqrt{3})=94.6$
Hence, the height of the tower is 94.6m.

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Question 35 Marks

An aeroplane is flying at a height of 300m above the ground. Flying at this height the angles of depression on from the two aeoplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. $\big[\text{Use}\sqrt{3}=1.732\big]$

Answer

Correct Figure,
$\tan 45^{\circ} = \frac{300}{\text{y}}$
$\Rightarrow 1 = \frac{300}{\text{y}} \text{ or } \text{y} = 300$
$\tan 60^{\circ} = \frac{300}{\text{x}}$
$\Rightarrow \sqrt{3} = \frac{300}{\text{x}} \text{ or } \text{x} = \frac{300}{\sqrt{3}} = 100\sqrt{3}$
Width of river $= 300 + 100 \sqrt{3} = 300 + 173.2$
$= 473.2 \text{ m}$

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Question 45 Marks

From the top of a building AB, 60m high, the an es of depression of the top and bottom of a vertical lamp-post CD are obseved to be 30° and 60° respetively. Find:

The horizontal distance between AB and CD,
The height of the lamp post,
The difference between the heights of the building and the lamp-post.

Answer

Given that AB is a bulding that is 60m, high.
Let BD = CE = x and CD = BE = y
$\Rightarrow\text{AE}=\text{AB}-\text{BE}=60-\text{y}$

In right $\triangle\text{ACE},$

$\tan30^\circ=\frac{\text{AE}}{\text{CE}}$

$\Rightarrow\frac{1}{\sqrt{3}}=\frac{60-\text{y}}{\text{x}}$

$\Rightarrow\text{x}=60\sqrt{3}-\text{y}\sqrt{3}\dots(\text{i})$

In right $\triangle\text{ACE},$

$\tan60^\circ=\frac{\text{AB}}{\text{BD}}$

$\Rightarrow\sqrt{3}=\frac{60}{\text{x}}$

$\Rightarrow\text{x}=\frac{60}{\sqrt{3}}$

On rationalising we get,

$\text{x}=\frac{60}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow\text{x}=\frac{60\sqrt{3}}{{3}}$

$\Rightarrow\text{x}={20\sqrt{3}}$

$\Rightarrow\text{x}=20\times1.732=34.64\text{m}$

Thus, the horizontal distance between AB and CD is 34.64m.

From (i), we get the height of the lamp-post

= CD = y

$\text{x}=60\sqrt{3}-\text{y}\sqrt{3}$

$\Rightarrow20\sqrt{3}=60\sqrt{3}=\text{y}\sqrt{3}$

$\Rightarrow20=60-\text{y}$

$\Rightarrow\text{y}=40\text{m}$

Thus, the height of the lamp-post is 40m.

The difference between of the building and the lamp-post

= AB - CD

= 60 - 40

= 20m

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Question 55 Marks

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° resbectively. If the bridge is at a height of 2.5m from the banks, find the width of the river.$\big[\text{Take}\sqrt{3}=1.732\big]$

Answer

Let the width of the river be AC.
D is the point on the bridge and BD = 5m
In right $\triangle\text{ABD},$
$\Rightarrow\tan45^\circ=\frac{\text{BD}}{\text{AB}}$
$\Rightarrow1=\frac{2.5}{\text{AB}}$
$\Rightarrow\text{AB}=2.5\text{m}$
In right $\triangle\text{CBD},$
$\tan30^\circ=\frac{\text{BD}}{\text{BC}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{2.5}{\text{BC}}$
$\Rightarrow\text{BC}=2.5\sqrt{3}\text{m}$
$\text{AC}=\text{AB}+\text{BC}$
$=2.5+2.5\sqrt{3}$
$=2.5(1+\sqrt{3})$
$=2.5(1+1.732)$
$=6.83\text{m}$
$\therefore$ Width of the river is 6.83m.

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Question 65 Marks

The horizontal distance between two towers is 60 metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 metres, find the height of the first tower. $\big[\text{Use}\sqrt{3}=1.732\big]$

Answer

Let AB and CD be the first and secound towers respectively.
Then, CD = 90m and AC = 60m.
Let DE be the horizontal line through D.

Draw $\text{BF}\perp\text{CD},$
Then, $\text{BF}=\text{AC}=60\text{m}$
$\angle\text{FBD}=\angle\text{EDB}=30^\circ$
Now, $\frac{\text{FD}}{\text{BF}}=\tan30^\circ-\frac{\text{FD}}{60}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{FD}=\Big(60\times\frac{1}{\sqrt{3}}\Big)\text{m}=20\sqrt{3}\text{m}$
$\therefore\text{AB}=\text{FC}=(\text{CD}-\text{FD})$
$=(90-20\sqrt{3})\text{m}=55.36\text{m}$

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