Chapter 9: Some Applications of Trigonometry

Ans:

Given: On the edge of a canal, a TV tower stands vertically. The angle of elevation of the top of the tower from a location on the other bank exactly opposite the tower is \[60^\circ \]. The angle of elevation of the top of the tower is \[30^\circ \] measured from another point $20\;\;{\text{m}}$ distant on the line connecting this point to the foot of the tower. 

To find: Calculate the tower's height and the canal's width.

In $\Delta ABC$

$\dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \tan {60^\circ }$

$\dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \sqrt 3 $

${\text{BC}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }}$

In $\vartriangle {\text{ABD}}$,

$\dfrac{{{\text{AB}}}}{{{\text{BD}}}} = \tan 30$

$\dfrac{{{\text{AB}}}}{{{\text{BC}} + {\text{CD}}}} = \dfrac{1}{{\sqrt 3 }}$

$\dfrac{{{\text{AB}}}}{{\dfrac{{{\text{AB}}}}{{\sqrt 3 }} + 20}} = \dfrac{1}{{\sqrt 3 }}$

$\dfrac{{{\text{AB}}\sqrt 3 }}{{{\text{AB}} + 20\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$

$3{\text{AB}} = {\text{AB}} + 20\sqrt 3 $

$2{\text{AB}} = 20\sqrt 3 $

${\text{AB}} = 10\sqrt 3 \;{\text{m}}$

${\text{BC}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }} = \left( {\dfrac{{10\sqrt 3 }}{{\sqrt 3 }}} \right){\text{m}} = 10\;{\text{m}}$

Therefore, the height of the tower is $10\sqrt 3 \;\;{\text{m}}$ and the width of the canal is $10\;\;{\text{m}}$.

Ans:

Given: From the top of a $7\;{\text{m}}$ high building, the angle of elevation of the top of a cable tower is \[60^\circ \] and the angle of depression of its foot is \[45^\circ \]. 

To find: Determine the height of the tower.

Let ${\text{AB}}$ be a building and ${\text{CD}}$ be a cable tower. In $\vartriangle {\text{ABD}}$,

$\dfrac{{{\text{AB}}}}{{{\text{BD}}}} = \tan 45$

$\dfrac{7}{{{\text{BD}}}} = 1 \Rightarrow {\text{BD}} = 7\;{\text{m}}$

In $\vartriangle {\text{ACE}}$,

${\text{AE}} = {\text{BD}} = 7\;{\text{m}}$

$\dfrac{{{\text{CE}}}}{{{\text{AE}}}} = \tan {60^\circ }$

$\dfrac{{{\text{CE}}}}{7} = \sqrt 3 $

${\text{CE}} = 7\sqrt 3 \;{\text{m}}$

${\text{CD}} = {\text{CE}} + {\text{ED}} = (7\sqrt 3  + 7){\text{m}}$

$ = 7(\sqrt 3  + 1){\text{m}}$

Therefore, the height of the cable tower is $(7\sqrt 3  + 7)\,{\text{m}}$.

Ans:

Given: The angles of depression of two ships are \[30^\circ \] and \[45^\circ \], respectively, as seen from the top of a $75\;{\text{m}}$ high lighthouse from sea level.

To find: Find the distance between the two ships if one is precisely behind the other on the same side of the lighthouse.

$\text { Let the distance between the ships be } \mathrm{x} \mathrm{m} \text {. }$

Then in right $\triangle \mathrm{ADB}, \tan 30^{\circ}=\dfrac{\mathrm{AB}}{\mathrm{BD}}$

$\Rightarrow \quad 1=\dfrac{75}{y} \quad \Rightarrow \quad y=75 \mathrm{~m}$

In right $\triangle \mathrm{ADB}, \tan 30^{\circ}=\dfrac{\mathrm{AB}}{\mathrm{BD}}$

$\Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{75}{x+y}$ $\Rightarrow x+75=75 \sqrt{3} \quad \quad [$ From equation (i) $] $

$\Rightarrow \quad x=75 \sqrt{3}-75=75(\sqrt{3}-1)$

Hence, the distance between the two ships is $\mathbf{7 5}(\sqrt{\mathbf{3}}-\mathbf{1}) \mathbf{m}$

Ans:

Given: A 1.2 m tall girl notices a balloon floating horizontally with the wind at a height of $88.2\;{\text{m}}$ above the earth. At any given time, the angle of elevation of the balloon from the girl's eyes is \[60^\circ \]. After a while, the angle of elevation drops to \[30^\circ \]

To find: Calculate the balloon's distance travelled throughout the timeframe.

Given that $1.2 \mathrm{~m}$ tall girl sees a balloon

So, $A G=1.2 \mathrm{~m}$

Also, AG \& BF are parallel $B F=A G=1.2 \mathrm{~m}$

And the balloon is at a height of $88.2 \mathrm{~m}$

So, EF $=88.2$

Girl sees balloon first at $60^{\circ}$

So, $\angle E A B=60^{\circ}$

After travelling, the angle of elevation becomes $30^{\circ}$

So, $\angle D A C=30^{\circ}$

Distance travelled by the balloon $=\mathrm{BC}$

We have to find BC

Now,

B E = E F - B F 

B E = 88.2 - 1.2 m

B E = 87 m

$\mathrm{Also}, \mathrm{BE}=\mathrm{DC}=87 \mathrm{~m}$

Here,

$\angle \mathrm{ABE}=90^{\circ} \& \angle \mathrm{ACD}=90^{\circ}$

In right angle triangle EBA

$\tan \mathrm{A}=\dfrac{\text { Side opposite to angle } \mathrm{A}}{\text { Side adjacent to angle } \mathrm{A}} $

$ tan \mathrm{A}=\dfrac{B E}{A B}$

$\tan 60^{\circ}=\dfrac{B E}{A B}$

$\sqrt{3}=\dfrac{87}{A B}$

$A B=\dfrac{87}{\sqrt{3}} \mathrm{~m}$

In right angle triangle DAC

$\tan \mathrm{A}=\dfrac{\text { Side opposite to angle } \mathrm{A}}{\text { Side adjacent to angle } \mathrm{A}} $

$\tan \mathrm{A}=\dfrac{C D}{A C}$

$\tan 30^{\circ}=\dfrac{C D}{A C}$

$\dfrac{1}{\sqrt{3}}=\dfrac{87}{A C}$

$\mathrm{AC}=87 \sqrt{3} \mathrm{~m}$

$A C=A B+B C$

$87 \sqrt{3}=\dfrac{87}{\sqrt{3}}+B C$

$87 \sqrt{3}-\dfrac{87}{\sqrt{3}}=B C$

$B C=87 \sqrt{3}-\dfrac{87}{\sqrt{3}}$

$B C=87\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right)$

$B C=87\left(\dfrac{\sqrt{3} \sqrt{3}-1}{\sqrt{3}}\right)$

$B C=87\left(\dfrac{3-1}{\sqrt{3}}\right)$

$B C=87\left(\dfrac{2}{\sqrt{3}}\right)$

$B C=\dfrac{87 \times 2}{\sqrt{3}}$

Multiply $\sqrt{3}$ in numerator and denominator

$\mathrm{BC}=\dfrac{87 \times 2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} $

$\mathrm{BC}=\dfrac{87 \times 2 \times \sqrt{3}}{3} $

$\mathrm{BC}=29 \times 2 \times \sqrt{3} $

$\mathrm{BC}=58 \sqrt{3}$

Hence, distance travelled by balloon $=58 \sqrt{3} \mathrm{~m}$

Answer:

Let the height of the tower DC be $h \mathrm{~m}$ and the speed of the car be $\mathrm{x} \mathrm{m} / \mathrm{s}$. Then the distance covered by the car in $6 \mathrm{~s}$ will be $6 \mathrm{x} \mathrm{m}$.

Also let the time taken by car to move from $B$ to $C$ be $t \mathrm{~s}$.

Then the distance BC will be $x t \mathrm{~m}$

In right $\triangle \mathrm{DCB}$,

$\tan 60^{\circ}=\dfrac{\mathrm{DC}}{\mathrm{BC}} \quad \Rightarrow \quad \sqrt{3}=\dfrac{h}{t x} $

$\Rightarrow \quad h=\sqrt{3} t x$

In right $\triangle \mathrm{ACD}$,

tan 30° = $ \frac{DC}{AC} $

$\Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{h}{6 x+t x} $ 

$\Rightarrow  6 x+t x =\sqrt{3} h $

 $\Rightarrow x(6+t) =\sqrt{3} \times \sqrt{3} t x $

$\Rightarrow  6+t =3 t \quad \Rightarrow \quad 2 t=6 $

$\Rightarrow  \quad \text { [From (i)] }$

$ \Rightarrow  t =3 \mathrm{~s} $

Hence, the required time taken by car is $\mathbf{3} \mathbf{~ s .}$