3 Marks Question

Question 13 Marks

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower

Answer

Let BC be the building, AB be the transmission tower, and D be the point on the ground.
In ΔBCD,
${BC \over CD }= tan 45º$
$\Rightarrow {20\over CD} = 1$
$\Rightarrow CD =20$
In ΔACD,
${AC\over CD} = tan 60º$
$\Rightarrow \frac{AB+BC}{CD} = \sqrt 3$
$\Rightarrow \frac{AB+20}{20} = \sqrt 3$
$\Rightarrow {AB+20}= 20\sqrt 3$
$\Rightarrow AB = 20({\sqrt 3 - 1}) \ m.$

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Question 23 Marks

A kite is flying at a height of $60$ m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^o$ Find the length of the string, assuming that there is no slack in the string.

Answer

In right triangle $ABC,$
$\sin 60 ^ { \circ } = \frac { A B } { A C }$
$\Rightarrow \frac { \sqrt { 3 } } { 2 } = \frac { 60 } { \mathrm { AC } }$
$A C = \frac { 120 } { \sqrt { 3 } }$
Multiplying $\sqrt3$ in both numerator and denominator,
$\mathrm { AC } = \frac { 120 } { \sqrt { 3 } } \times \frac { \sqrt { 3 } } { \sqrt { 3 } }$
AC = $40\sqrt3$m
Hence the length of the string is $40\sqrt3$m.

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Question 33 Marks

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^o$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 \ m$. Find the height of the tree.

Answer

Let AC be the broken part of the tree.
$\therefore$ Total height of the tree = AB + AC
In right $\triangle$ABC,
cos $30^o $= $ \frac{BC}{AC}$
$\Rightarrow$ $\frac{\sqrt3}{2}$ = 8/AC$\frac{\sqrt3}{2}=\frac{8}{AC}$
$\Rightarrow$ AC = $\frac{16}{\sqrt3}$
Also,
tan $30^o $= $\frac{AB}{BC}$
$\Rightarrow$ $\frac{1}{\sqrt3}=\frac{AB}{8}$
$\Rightarrow$ AB = $\frac{8}{\sqrt3}$
Total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3$\frac{16}{\sqrt3}$+$\frac{8}{\sqrt3}$ = $\frac{24}{\sqrt3}$

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Question 43 Marks

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the further time taken by the car to reach the foot of the tower from this point.

Answer

In right triangle ABP,
$\tan 30 ^ { \circ } = \frac { \mathrm { AB } } { \mathrm { BP } }$
$\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { \mathrm { AB } } { \mathrm { BP } }$
BP = AB$\sqrt3$ ........ (i)
In right triangle ABQ,
$tan\;60^0\;={AB\over BQ}$
$\Rightarrow \sqrt { 3 } = \frac { A B } { B Q }$
$\Rightarrow _ { B Q } = \frac { A B } { \sqrt { 3 } }$....... (ii)
$\because$ PQ = BP - BQ
$\therefore$ PQ = AB$\sqrt { 3 } - \frac { A B } { \sqrt { 3 } } = \frac { 3 A B - A B } { \sqrt { 3 } } = \frac { 2 A B } { \sqrt { 3 } }$ = 2BQ [From eq. (ii)]
$\Rightarrow$ BQ = $\frac12$PQ
$\because$ Time taken by the car to travel a distance PQ = 6 seconds.
$\therefore$ Time taken by the car to travel a distance BQ, i.e. $\frac12$PQ = $\frac12$ $\times$ 6 = 3 seconds.
Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.

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Question 53 Marks

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance traveled by the balloon during the interval.

Answer

In right triangle ABC,
$\tan 60 ^ { \circ } = \frac { A B } { B C }$
$\Rightarrow \sqrt { 3 } = \frac { 88.2 } { B C }$
$\Rightarrow B C = \frac { 88.2 } { \sqrt { 3 } }$

In right triangle PQC,
$\tan 30 ^ { \circ } = \frac { \mathrm { PQ } } { \mathrm { CO } }$
$\Rightarrow \tan 30 ^ { \circ } = \frac { \mathrm { PQ } } { \mathrm { CB } + \mathrm { BQ } } \Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { 88.2 } { \frac { 88.2 } { \sqrt { 3 } } + \mathrm { BQ } }$....... From (1)
$\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { 88.2 \sqrt { 3 } } { 88.2 + \mathrm { BQ } \sqrt { 3 } } \Rightarrow 88.2 + \mathrm { BQ } \sqrt { 3 } = 264.6$
$\Rightarrow \mathrm { BQ } \sqrt { 3 } = 264.6 - 88.2 = 176.4$
$\Rightarrow B Q = \frac { 176.4 } { \sqrt { 3 } } = \frac { ( 176.4 ) \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } }$
$= \frac { ( 176.4 ) \sqrt { 3 } } { 3 } = ( 58.8 ) \sqrt { 3 } = \frac { 58.8 } { 10 } \sqrt { 3 } = \frac { 294 } { 5 } \sqrt { 3 }$
Hence, the distance travelled by the balloon during the interval is $\frac { 294 } { 5 } \sqrt { 3 } \mathrm { m }$.

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Question 63 Marks

As observed from the top of a $75 \ m$ high lighthouse from the sea-level, the angles of depression of two ships are $30^o$ and $45^o$.If one ship is exactly behind the other on the same side of the lighthouse, find the distance between two ships.

Answer

In right triangle ABQ,

$\tan 45 ^ { \circ } = \frac { A B } { B Q }$
$\Rightarrow 1 = \frac { 75 } { \mathrm { BQ } }$
$\Rightarrow$BQ = 75 m ........ (i)
In right triangle ABP,
$\tan 30 ^ { \circ } = \frac { A B } { B P }$
$\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { A B } { B Q + Q P } $
$\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { A B } { 75 + Q P }$[From eq. (i)]
$\Rightarrow$ 75 + QP = $75\sqrt3$
QP = $75(\sqrt3-1)$ m
Hence the distance between the two ships is $75(\sqrt3-1)$ m.

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Question 73 Marks

From the top of a $7 \ m$ high building, the angle of elevation of the top of a cable tower is $60^o$ and the angle of depression of its foot is $45^o$. Determine the height of the tower.

Answer

In right triangle $ABD$,
$\tan {45^0} = \frac{{AB}}{{BD}}$

$ \Rightarrow $ $1 = \frac { 7 } { B D }$
$ \Rightarrow $$ BD = 7 m$
$ \Rightarrow $$ AE = 7 m$
In right triangle AEC,
$\tan 60 ^ { \circ } = \frac { C E } { A E }$
$\Rightarrow \sqrt { 3 } = \frac { \mathrm { CE } } { 7 } \Rightarrow \mathrm { CE } = 7 \sqrt { 3 } \mathrm { m }$
$\therefore$ $CD = CE + ED$
$= CE + AB$
$= 7 \sqrt { 3 } + 7 = 7 ( \sqrt { 3 } + 1 ) { \mathrm { m } }$
Hence height of the tower is $7 ( \sqrt { 3 } + 1 )$ m.

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Question 83 Marks

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Answer

Self-learning

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Question 93 Marks

The shadow of a tower standing on a level ground is found to be $40 \ m$ longer when the Sun's altitude is $30^o,$ than when it is $60^o.$ Find the height of the tower.

Answer

In $\triangle A B C,$ $\tan 60 ^ { \circ } = \frac { A B } { B C }$
$\Rightarrow \quad A B = \sqrt { 3 } B C$ .....(i)
In $\triangle A B D,$
$\tan 30 ^ { \circ } = \frac { A B } { B C + 40 }$
$\frac{ 1}{√3}=\frac{AB}{BC+40}=\frac{√3BC}{BC+40}$
3BC = BC + 40
BC = 20, Hence from (i) we get
AB = 20√3 = 20 $\times$ 1.73 = 34.6 meter

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Question 103 Marks

QUESTION From a point P on the ground the angle of elevation of the top of a $10 \ m$ tall building is $30^{\circ}$. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from $P$ is $45^{\circ}$. Find the length of the flagstaff and distance of building from point P. $\sqrt{3}=1.732]$

Answer

Let height of flagstaff be BD = x m
$\therefore \quad \tan 30 ^ { \circ } = \frac { A B } { A P }$
$\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { 10 } { A P }$
$A P = 10 \sqrt { 3 }$
distance of the building from P.
$= 10 \times 1.732 = 17.32 \mathrm { m }$

$\tan 45 ^ { \circ } = \frac { A D } { A P } \text { or } 1 = \frac { 10 + x } { 17 \cdot 32 }$
or, $x + 10 = 17.32$
$x = 17.32 - 10$
$x = 7.32$
Hence, length of flagstaff x = 1 7.32 m.

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Question 113 Marks

An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. Find the height of the chimney?

Answer

Here, AB is the chimney, CD the observer and $\angle$ADE the angle of elevation (see Fig). In this case, ADE is a triangle, right-angled at E and we are required to find the height of the chimney.

We have AB = AE + BE = AE + 1.5
and DE = CB = 28.5 m
To determine AE, we choose a trigonometric ratio, which involves both AE and DE. Let us choose the tangent of the angle of elevation.
Now, tan 45° = $\frac{\mathrm{AE}}{\mathrm{DE}}$
i.e., 1 = $\frac{\mathrm{AE}}{28.5}$
Therefore, AE = 28.5
So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m

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Question 123 Marks

A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Answer

First, let us draw a simple diagram to represent the problem. Here AB represents the tower, CB is the distance of the point from foot of the tower and $\angle$ ACB is the angle of elevation. We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angled at B. To solve the problem, we choose the trigonometric ratio tan 60° (or cot 60°), as the ratio involves AB and BC.
Now, tan 60° =$\frac{\mathrm{AB}}{\mathrm{BC}}$
i.e., $\sqrt{3}=\frac{\mathrm{AB}}{15}$
i.e., AB = $15 \sqrt{3}$
Hence, the height of the tower is $15 \sqrt{3}$ m

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