4 Marks Questions

Question 14 Marks

The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ $. If the tower is $50 \ m$ high, find the height of the building.

Answer

Given,

Let the height of building be $AB$ and height of tower $CD$
Height of the tower $(CD)= 50 \ m$
Angle of elevation of top of building from foot of tower $= 30^\circ$
Hence, $\angle ACB = 30^\circ $
Angle of elevation of top of tower from from foot of building $= 60^\circ$
Hence,$\angle DBC = 60^\circ$
$\angle ABC = 90^\circ \angle DCB = 90^\circ$
In a right angle $\triangle DBC,$
$\tan B = \frac{side \ opposite \ to \ angle \ to \ D} {side \ opposite \ to \ angle \ H}$
$\tan B = \frac{D C}{B C}$
$\tan 60^0 = \frac{50}{B C}$
$B C=\frac{50}{\sqrt{3}}$
Similarly,
In a right angle triangle $ABC,$
$\tan C = \frac{side \ opposite \ to \ angle \ to \ C} {side \ adjacent \ to \ angle \ C}$
$\tan 30^\circ = \frac{AB} {BC}$
$\frac{1}{\sqrt{3}}=\frac{A B}{\frac{50}{\sqrt{3}}}$
$\frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}}=\mathrm{AB}$
$A B=\frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}}$
$A B=\frac{50}{3} m$

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Question 24 Marks

A statue, $1.6 \ m$ tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ.$ Find the height of the pedestal.

Answer

Let the height of the pedestal be $h \ m.$
$\therefore BC = h \ m$
In right triangle $ACP,$
$\tan 60 ^ { \circ } = \frac { A C } { P C }$
$\Rightarrow \sqrt { 3 } = \frac { A B + B C } { P C }$
$\Rightarrow \sqrt { 3 } = \frac { 1.6 + h } { \mathrm { PC } }.....(i)$
In right triangle $BCP,$
$\tan 45 ^ { \circ } = \frac { B C } { P C }$
$\Rightarrow 1 = \frac { h } { \mathrm { PC } } \Rightarrow _ { \mathrm { PC } } = h$
$\therefore \sqrt { 3 } = \frac { 1.6 + h } { h }[$From eq. $(i)]$
$\Rightarrow \sqrt { 3 } h = 1.6 + h \Rightarrow h ( \sqrt { 3 } - 1 ) = 1.6 \Rightarrow \frac { 1.6 } { \sqrt { 3 } - 1 }$
$\Rightarrow \frac { 1.6 ( \sqrt { 3 } + 1 ) } { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) } \Rightarrow { h = \frac { 1.6 ( \sqrt { 3 } + 1 ) } { 3 - 1 } } \Rightarrow h = \frac { 1.6 ( \sqrt { 3 } + 1 ) } { 2 }$
$\Rightarrow h = 0.8 ( \sqrt { 3 } + 1 ) _ { \mathrm { m } }$

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Question 34 Marks

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

Let AE is the Length of the building. So AE = 30 Again BE = DF = 1.5 AB = AE - BE = 30 - 1.5 = 28.5 Now in triangle ABC, tan60 = $\frac{AB}{BC}$ $\Rightarrow $ √3 = $\frac{28.5}{BC}$ $\Rightarrow $ BC = $\frac{28.5}{\sqrt 3}$ Again in triangle ABD tan30 = $\frac{AB}{BD}$
$\sqrt{1}{\sqrt3}=\frac{28.5}{BD}$ $\Rightarrow $ BD = 28.5$\times$√3 $\Rightarrow $ BC + CD = 28.5√3 $\Rightarrow $ 28.5/√3 + CD = 28.5√3 $\Rightarrow $ CD = $\frac{28.5}{\sqrt 3}$- $\frac{28.5}{\sqrt 3}$ $\Rightarrow $ CD = $\frac{28.5\times3-28.5}{\sqrt3}$ $\Rightarrow $ CD = $\frac{28.5(3-1)}{\sqrt3}$ $\Rightarrow $ CD = $\frac{(28.5\times 2)}{\sqrt3}$ $\Rightarrow $ CD = $\frac{(57)}{\sqrt3}$ $\Rightarrow $ CD = $\frac{(57\sqrt3)}{\sqrt3\times\sqrt3}$ (Multiply √3 in numerator and denominator) $\Rightarrow $ CD = $\frac{57\sqrt3}{3}$ $\Rightarrow $ CD = 19√3 The distance he walked towards the building is 19√3 m

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Question 44 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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