Chapter 9: Some Applications of Trigonometry

Ans:

Given: A circus performer is climbing a $20\;{\text{m}}$ rope that runs from the top of a vertical pole to the ground, firmly stretched and knotted. 

To find: If the rope makes a \[30^\circ \] angle with the ground level, calculate the height of the pole.

It can be observed from the figure that AB is the pole. 

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

$\dfrac{{{\text{AB}}}}{{{\text{AC}}}} = \sin {30^\circ }$

$\dfrac{{{\text{AB}}}}{{20}} = \dfrac{1}{2}$

${\text{AB}} = \dfrac{{20}}{2} = 10$

Therefore, the height of the pole is $10\;\;{\text{m}}$.

Ans:

Given: A storm causes a tree to shatter, and the fractured section bends such that the top of the tree meets the ground at a \[30^\circ \] angle. The distance from the tree's foot to where the top touches the earth is $8$ meters. 

To find: Determine the tree's height.

The original tree was Let AC. It was split in two due to the storm. The broken portion A'B is at a \[30^\circ \]angle to the ground.

In $\Delta {{\text{A}}^\prime }{\text{BC}}$

$\dfrac{{{\text{BC}}}}{{{{\text{A}}^\prime }{\text{C}}}} = \tan {30^\circ }$

$\dfrac{{{\text{BC}}}}{8} = \dfrac{1}{{\sqrt 3 }}$

${\text{BC}} = \left( {\dfrac{8}{{\sqrt 3 }}} \right){\text{m}}$

$\dfrac{{{{\text{A}}^\prime }{\text{C}}}}{{{{\text{A}}^\prime }{\text{B}}}} = \cos {30^\circ }$

$\dfrac{8}{{\;{{\text{A}}^\prime }{\text{B}}}} = \dfrac{{\sqrt 3 }}{2}$

$\;{{\text{A}}^\prime }{\text{B}} = \left( {\dfrac{{16}}{{\sqrt 3 }}} \right){\text{m}}$

Height of tree $ = {{\text{A}}^\prime }{\text{B}} + {\text{BC}}$

$ = \left( {\dfrac{{16}}{{\sqrt 3 }} + \dfrac{8}{{\sqrt 3 }}} \right){\text{m}} = \dfrac{{24}}{{\sqrt 3 }}\;{\text{m}}$

$ = 8\sqrt 3 \;{\text{m}}$

Ans:

Given: In a park, a contractor intends to construct two slides for the kids to enjoy. She likes a slide with a top height of $1.5\,{\text{m}}$ and an angle of \[30^\circ \] to the ground for children under the age of $5$, and a steep slide with a top height of $3\;{\text{m}}$ and an angle of \[60^\circ \] to the ground for older children. 

To find: In each scenario, what should the length of the slide be?

It can be noted that AC and PR are the slides for younger and elder children.

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

$\dfrac{{{\text{AB}}}}{{{\text{AC}}}} = \sin {30^\circ }$

$\dfrac{{1.5}}{{{\text{AC}}}} = \dfrac{1}{2}$

${\text{AC}} = 3\;{\text{m}}$

In $\Delta {\text{PQR}}$,

$\dfrac{{{\text{PQ}}}}{{{\text{PR}}}} = \sin 60$

$\dfrac{3}{{PR}} = \dfrac{{\sqrt 3 }}{2}$

${\text{PR}} = \dfrac{6}{{\sqrt 3 }} = 2\sqrt 3 \;{\text{m}}$

Therefore, the lengths of these slides are $3\;{\text{m}}$ and $2\sqrt 3 \;{\text{m}}$.

Ans:

Given: The angle of elevation of a tower's top from a point on the ground $30$ metres distant from the tower's foot is \[30^\circ \]. 

To find: Determine the tower's height.

Let AB be the tower and the angle of elevation from point C (on ground) is \[30^\circ \]. In \[\Delta ABC\],

In \[\Delta ABC\], $\dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \tan {30^\circ }$

$\dfrac{{{\text{AB}}}}{{30}} = \dfrac{1}{{\sqrt 3 }}$

${\text{AB}} = \dfrac{{30}}{{\sqrt 3 }} = 10\sqrt 3 \;{\text{m}}$

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

Ans:

Given: At a height of $60\;{\text{m}}$ above the earth, a kite is flying. The kite's line is temporarily tethered to a location on the ground. The string is at a \[60^\circ \]angle to the ground. 

To find: Determine the string's length, assuming there is no slack in the string.

Let K be the kite and the string is tied to point P on the ground. In \[\Delta KLP\],

In $\Delta {\text{KLP}}$,

$\dfrac{{{\text{KL}}}}{{{\text{KP}}}} = \sin {60^\circ }$

$\dfrac{{60}}{{{\text{KP}}}} = \dfrac{{\sqrt 3 }}{2}$

${\text{KP}} = \dfrac{{120}}{{\sqrt 3 }} = 40\sqrt 3 \;{\text{m}}$

Hence, the length of the string is $40\sqrt 3 \;{\text{m}}$.

Ans:

Given: A 1.5-meter-tall child stands at a safe distance from a $30\,{\text{m}}$ tall structure. As he goes towards the structure, the angle of elevation from his eyes to the top of the building grows from \[30^\circ \] to \[60^\circ \]. 

To find: Calculate how far he went towards the building.

Let the boy was standing at point S initially. He walked towards the building and reached point T.

${\text{PR}} = {\text{PQ}} - {\text{RQ}}$

$ = (30 - 1.5){\text{m}} = 28.5\;{\text{m}} = \dfrac{{57}}{2}\;{\text{m}}$

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

$\dfrac{{{\text{PR}}}}{{{\text{AR}}}} = \tan {30^\circ }$

$\dfrac{{57}}{{2{\text{AR}}}} = \dfrac{1}{{\sqrt 3 }}$

${\text{AR}} = \left( {\dfrac{{57}}{2}\sqrt 3 } \right){\text{m}}$

In $\vartriangle {\text{PRB}}$

$\dfrac{{{\text{PR}}}}{{{\text{BR}}}} = \tan 60$

$\dfrac{{57}}{{2{\text{BR}}}} = \sqrt 3 $

${\text{BR}} = \dfrac{{57}}{{2\sqrt 3 }} = \left( {\dfrac{{19\sqrt 3 }}{2}} \right){\text{m}}$

${\text{ST}} = {\text{AB}}$

$ = {\text{AR}} - {\text{BR}} = \left( {\dfrac{{57\sqrt 3 }}{2} - \dfrac{{19\sqrt 3 }}{2}} \right){\text{m}}$

$ = \left( {\dfrac{{38\sqrt 3 }}{2}} \right){\text{m}} = 19\sqrt 3 \;{\text{m}}$

Hence, he walked $19\sqrt 3 \;{\text{m}}$ towards the building.

Ans:

Given: The angles of elevation of the bottom and top of a transmission tower installed at the top of a $20\,{\text{m}}$ high structure are \[45^\circ \] and \[60^\circ \], respectively, from a point on the ground. 

To find: Determine the tower's height.

Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

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

$\dfrac{{{\text{BC}}}}{{{\text{CD}}}} = \tan {45^\circ }$

$\dfrac{{20}}{{{\text{CD}}}} = 1$

${\text{CD}} = 20\;{\text{m}}$

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

$\dfrac{{{\text{AC}}}}{{{\text{CD}}}} = \tan {60^\circ }$

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

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

${\text{AB}} = (20\sqrt 3  - 20){\text{m}}$

$ = 20(\sqrt 3  - 1){\text{m}}$

Therefore, the height of the transmission tower is $20(\sqrt 3  - 1){\text{m}}$

Ans:

Given: A $1.6\,{\text{m}}$ tall statue sits on top of a pedestal; the angle of elevation of the top of the statue is \[60^\circ \] from a point on the ground, and the angle of elevation of the top of the pedestal is \[45^\circ \] from the same point. 

To find: Determine the pedestal's height.

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

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

$\dfrac{{{\text{BC}}}}{{{\text{CD}}}} = \tan {45^\circ }$

$\dfrac{{BC}}{{CD}} = 1$

${\text{BC}} = {\text{CD}}$

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

$\dfrac{{{\text{AB}} + {\text{BC}}}}{{{\text{CD}}}} = \tan 60$

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

$1.6 + {\text{BC}} = {\text{BC}}\sqrt 3 \quad [{\text{AsCD}} = {\text{BC}}]$

${\text{BC}}(\sqrt 3  - 1) = 1.6$

${\text{BC}} = \dfrac{{(1.6)(\sqrt 3  + 1)}}{{(\sqrt 3  - 1)(\sqrt 3  + 1)}}$

(By Rationalization)

$ = \dfrac{{1.6(\sqrt 3  + 1)}}{{{{(\sqrt 3 )}^2} - {{(1)}^2}}}$

$ = \dfrac{{1.6(\sqrt 3  + 1)}}{2} = 0.8(\sqrt 3  + 1)$

Therefore, the height of the pedestal is $0.8(\sqrt 3  + 1)\;{\text{m}}$

Answer :

Given: The angle of elevation of a building's top from the foot of the tower is \[30^\circ \], while the angle of elevation of a tower from the foot of the building is \[60^\circ \]. 

To find: Find the height of the building if the tower is $50{\text{ m}}$ tall.

Let AB be the building and CD be the tower. 

In $\Delta {\text{CDB}}$,

$\dfrac{{{\text{CD}}}}{{{\text{BD}}}} = \tan {60^\circ }$

$\dfrac{{50}}{{{\text{BD}}}} = \sqrt 3 $

${\text{BD}} = \dfrac{{50}}{{\sqrt 3 }}$

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

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

${\text{AB}} = \dfrac{{50}}{{\sqrt 3 }} \times \dfrac{1}{{\sqrt 3 }} = \dfrac{{50}}{3} = 16\dfrac{2}{3}$

Therefore, the height of the building is $16\dfrac{2}{3}\;{\text{m}}$.

Ans:

Given: On either side of the $80{\text{ m}}$ wide road, two poles of equal height stand opposite each other. The angles of elevation of the tops of the poles are \[60^\circ \] and \[30^\circ \], respectively, from a point on the route between them. 

To find: Determine the height of the poles and the distance between them.

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

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

$\dfrac{{{\text{AB}}}}{{{\text{BO}}}} = \tan 60$

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

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

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

$\dfrac{{CD}}{{DO}} = \tan {30^\circ }$

$\dfrac{{{\text{CD}}}}{{80 - {\text{BO}}}} = \dfrac{1}{{\sqrt 3 }}$

${\text{CD}}\sqrt 3  = 80 - {\text{BO}}$

$CD\sqrt 3  = 80 - \dfrac{{AB}}{{\sqrt 3 }}$

${\text{CD}}\sqrt 3  + \dfrac{{{\text{AB}}}}{{\sqrt 3 }} = 80$

Since the poles are of equal heights, ${\text{CD}} = {\text{AB}}$

${\text{CD}}\left[ {\sqrt 3  + \dfrac{1}{{\sqrt 3 }}} \right] = 80$

${\text{CD}}\left( {\dfrac{{3 + 1}}{{\sqrt 3 }}} \right) = 80$

${\text{CD}} = 20\sqrt 3 \;{\text{m}}$

${\text{BO}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }} = \dfrac{{{\text{CD}}}}{{\sqrt 3 }} = \left( {\dfrac{{20\sqrt 3 }}{{\sqrt 3 }}} \right){\text{m}} = 20\;{\text{m}}$

${\text{DO}} = {\text{BD}} - {\text{BO}} = (80 - 20){\text{m}} = 60\;{\text{m}}$

Therefore, the height of poles is $20\sqrt 3 $ and the point is $20\;\;{\text{m}}$ and $60\;\;{\text{m}}$ far from the D