NCERT Solutions Class 10 Maths Chapter 9

Q.1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see the following figure).


Ans.

\begin{array}{l}\text{In the given figure above, we have}\\ \text{sin}\text{30°}=\frac{\text{AB}}{\text{AC}}=\frac{\text{AB}}{20}\\ \text{or}\frac{1}{2}=\frac{\text{AB}}{20}\\ \text{or AB}=\frac{20}{2}=10\\ \text{Therefore, height of the pole is 10 m}\text{.}\end{array}

Q.2 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° 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.

Ans.

Let the tree AB bends at point C. Foot of the tree is at A and CB’ is the broken part of the tree i.e, BC = CB’.

$\begin{array}{l}\text{Now, in the above figure, we have}\\ \tan \text{\hspace{0.17em}30°}=\frac{\text{AC}}{\text{AB’}}=\frac{\text{AC}}{8}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{AC}}{8}\\ \text{or AC}=\frac{8}{\sqrt{3}}\\ \text{Also,}\\ \text{cos\hspace{0.17em}30°}=\frac{\text{AB’}}{\text{CB’}}=\frac{\text{8}}{\text{CB’}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{\text{8}}{\text{CB’}}\\ \text{or CB’}=\frac{16}{\sqrt{3}}\\ \text{Therefore, height of the tree}=\text{AC}+\text{CB’}=\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8\sqrt{3}\text{\hspace{0.17em}\hspace{0.17em}m}\end{array}$

Q.3 A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Ans.

As per given information, we have the following figure.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}30°}=\frac{\text{AB}}{\text{AC}}=\frac{1.5}{\text{AC}}\\ \text{or}\frac{1}{2}=\frac{1.5}{\text{AC}}\\ \text{or AC}=1.5\times 2=3\text{m}\\ \text{Therefore, length of the slide for the children below}\\ \text{the age of 5 years is}3\text{m.}\end{array}$

Also, for elder children, we have the following slide.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{PQR, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}}6\text{0°}=\frac{\text{PQ}}{\text{PR}}=\frac{3}{\text{PR}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{3}{\text{PR}}\\ \text{or PR}=\frac{3\times 2}{\sqrt{3}}=2\sqrt{3}\text{m}\\ \text{Therefore, length of the slide for the elder children is}2\sqrt{3}\text{m.}\end{array}$

Q.4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Ans.

Let PQ is a tower with foot at Q and R is a point on the ground such that distance between Q and R is 30 m.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{PQR, we have}\\ \tan \text{\hspace{0.17em}30°}=\frac{\text{PQ}}{\text{QR}}=\frac{\text{PQ}}{\text{30}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{PQ}}{\text{30}}\\ \text{or PQ}=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{m}\\ \text{Therefore, height of the tower is\hspace{0.33em}}10\sqrt{3}\text{m}.\end{array}$

Q.5 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°. Find the length of the string, assuming that there is no slack in the string.

Ans.

Let the kite is flying at point A and AC is the string of the kite tied at point C on the ground.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}}6\text{0°}=\frac{\text{AB}}{\text{AC}}=\frac{60}{\text{AC}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{60}{\text{AC}}\\ \text{or AC}=\frac{60\times 2}{\sqrt{3}}=40\sqrt{3}\text{m}\\ \text{Therefore, length of the string is}40\sqrt{3}\text{m.}\end{array}$

Q.6 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.

Ans.

Let AB be a building of height 30 m. Let the boy is initially at D and then from D moves to point E. We have to find the distance DE

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ACE, we have}\\ \tan 6\text{0°}=\frac{\text{AC}}{\text{EC}}=\frac{28.5}{\text{EC}}\\ \text{or}\sqrt{3}=\frac{28.5}{\text{EC}}\\ \text{or EC}=\frac{28.5}{\sqrt{3}}\text{m}\\ \text{Now,}\\ \text{In}\mathrm{\Delta }\text{ACD, we have}\\ \tan 3\text{0°}=\frac{\text{AC}}{\text{CD}}=\frac{28.5}{\text{EC}+\text{DE}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{28.5}{\text{EC}+\text{DE}}\\ \text{or EC}+\text{DE}=28.5\sqrt{3}\\ \text{or DE}=28.5\sqrt{3}-\text{EC}=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}\\ \text{or DE}=\frac{28.5\times 3-28.5}{\sqrt{3}}=\frac{57}{\sqrt{3}}=\frac{19\times 3}{\sqrt{3}}=19\sqrt{3}\text{m}\\ \text{Therefore, the required distance is}19\sqrt{3}\text{m.}\end{array}$

Q.7 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.

Ans.

Let BC be a building on which a transmission tower AB is fixed. Let D be a point on the ground.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}45\text{°}=\frac{\text{BC}}{\text{DC}}=\frac{20}{\text{DC}}\\ \text{or}1=\frac{20}{\text{DC}}\\ \text{or DC}=20\text{m}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}ACD, we have}\\ \text{tan\hspace{0.17em}}60\text{°}=\frac{\text{AC}}{\text{DC}}=\frac{\text{AC}}{20}\\ \text{or}\sqrt{3}=\frac{\text{AC}}{20}\\ \text{or AC}=20\sqrt{3}\\ \text{or AC}=\text{AB}+\text{BC}=20\sqrt{3}\\ \text{or AB}=20\sqrt{3}-\text{BC}=20\sqrt{3}-\text{20}=20(\sqrt{3}-1)\\ \text{Therefore, height of the tower is}20(\sqrt{3}-1)\text{m.}\end{array}$

Q.8 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° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Ans.

Let AB be the statue on the pedestal BC. Let D be a point on the ground.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}45\text{°}=\frac{\text{BC}}{\text{DC}}\\ \text{or}1=\frac{\text{BC}}{\text{DC}}\\ \text{or DC}=\text{BC}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}ACD, we have}\\ \text{tan\hspace{0.17em}}60\text{°}=\frac{\text{AC}}{\text{DC}}=\frac{\text{AB}+\text{BC}}{\text{BC}}\\ \text{or}\sqrt{3}=\frac{\text{1.6}+\text{BC}}{\text{BC}}\\ \text{or}\sqrt{3}\text{\hspace{0.17em}BC}=\text{1.6}+\text{BC}\\ \text{or BC}(\sqrt{3}-1)=\text{1.6}\\ \text{or BC}=\frac{\text{1.6}}{(\sqrt{3}-1)}=\frac{\text{1.6}}{(\sqrt{3}-1)}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=0.8(\sqrt{3}+1)\text{m}\\ \text{Therefore, height of the pedestal is}0.8(\sqrt{3}+1)\text{m.}\end{array}$

Q.9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Ans.

Let AB be the height of the building and CD be the height of the tower.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60\text{°}=\frac{\text{CD}}{\text{BD}}\\ \text{or}\sqrt{3}=\frac{\text{50}}{\text{BD}}\\ \text{or BD}=\frac{\text{50}}{\sqrt{3}}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}ABC, we have}\\ \text{tan\hspace{0.17em}}30\text{°}=\frac{\text{AB}}{\text{BD}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{AB}}{\frac{\text{50}}{\sqrt{3}}}\\ \text{or AB}=\frac{1}{\sqrt{3}}\times \frac{\text{50}}{\sqrt{3}}=\frac{\text{50}}{3}=16\frac{2}{3}\\ \text{Therefore, height of the building is}16\frac{2}{3}\text{m.}\end{array}$

Q.10 Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Ans.

Let AB and CD be the two poles of equal height. Let O be a point on the road from where elevation angles are measured.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABO, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60\text{°}=\frac{\text{AB}}{\text{BO}}\\ \text{or}\sqrt{3}=\frac{\text{AB}}{\text{BO}}\\ \text{or BO}=\frac{\text{AB}}{\sqrt{3}}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}CDO, we have}\\ \text{tan\hspace{0.17em}}30\text{°}=\frac{\text{CD}}{\text{OD}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{CD}}{\text{OD}}=\frac{\text{CD}}{\text{BD}-\text{BO}}=\frac{\text{CD}}{\text{80}-\text{BO}}\\ \text{or CD}\sqrt{3}=\text{80}-\text{BO}=\text{80}-\frac{\text{AB}}{\sqrt{3}}\\ \text{or AB}\sqrt{3}=\text{80}-\frac{\text{AB}}{\sqrt{3}}\\ [\text{AB}=\text{CD as heights of the poles are equal]}\\ \text{or AB}\sqrt{3}+\frac{\text{AB}}{\sqrt{3}}=\text{80}\\ \text{or}\frac{4\text{AB}}{\sqrt{3}}=80\\ \text{or AB}=20\sqrt{3}\\ \text{Therefore, height of each pole is}20\sqrt{3}\text{m.}\\ \text{BO}=\frac{\text{AB}}{\sqrt{3}}=\frac{20\sqrt{3}}{\sqrt{3}}=20\text{m}\\ \text{OD}=\text{BD}-\text{BO}=80-20=60\text{m}\\ \text{Therefore, the required distances are 20 m and 60 m.}\end{array}$

Q.11 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 joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the following figure). Find the height of the tower and the width of the canal.


Ans.

We have the following figure.

$\begin{array}{l}\text{In ΔABD, we have}\\ \text{tan\hspace{0.17em}30° =}\frac{\text{AB}}{\text{DB}}\text{=}\frac{\text{AB}}{\text{BC+CD}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{AB}}{\text{BC+20}}\\ \text{or BC+20 = AB}\sqrt{\text{3}}\\ \text{or BC = AB}\sqrt{\text{3}}-\text{20\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° =}\frac{\text{AB}}{\text{BC}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AB}}{\text{BC}}\\ \text{or AB = BC}\sqrt{\text{3}}...\text{(2)}\\ \text{From equations (1) and (2), we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC = AB}\sqrt{\text{3}}-\text{20}\\ \text{or BC = BC}\sqrt{\text{3}}\text{\hspace{0.17em}×}\sqrt{\text{3}}-\text{20=3BC-20}\\ \text{or 3BC}-\text{BC = 20}\\ \text{or 2BC = 20}\\ \text{or BC = 10 m}\\ \text{On putting this value of BC in equation (2), we get}\\ \text{AB = 10}\sqrt{\text{3}}\text{m}\\ \text{Therefore, height of the tower is 10}\sqrt{\text{3}}\text{m and width of}\\ \text{the canal is 10 m.}\end{array}$

Q.12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Ans.

Let AB is a cable tower and CD is a building.

$\begin{array}{l}\text{We have, CD\hspace{0.33em}=\hspace{0.33em}BE\hspace{0.33em}=\hspace{0.33em}7 m, DB\hspace{0.33em}=\hspace{0.33em}CE}\\ \text{In Δ\hspace{0.17em}CDB, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}45°\hspace{0.33em}=\hspace{0.33em}}\frac{\text{CD}}{\text{DB}}\\ \text{or 1\hspace{0.33em}=\hspace{0.33em}}\frac{\text{7}}{\text{DB}}\\ \text{or DB =\hspace{0.33em}7 m}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60°\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{CE}}\\ \text{or AE\hspace{0.33em}= CE}\sqrt{\text{3}}\text{\hspace{0.33em}= DB}\sqrt{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}7}\sqrt{\text{3}}\text{m\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}[CE\hspace{0.17em}= DB]}\\ \text{Therefore,}\\ \text{height of the tower\hspace{0.33em}=\hspace{0.33em}AB =\hspace{0.33em}AE+BE =\hspace{0.33em}7}\sqrt{\text{3}}\text{+7 =\hspace{0.33em}7(1+}\sqrt{\text{3}}\text{) m}\end{array}$

Q.13 As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Ans.

$\begin{array}{l}\text{Let AB be a lighthouse and ships be at points C and D.}\\ \text{It is given that AB = 75 m. We have to find the distance CD.}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}45° =}\frac{\text{AB}}{\text{BC}}\\ \text{or 1 =}\frac{\text{AB}}{\text{BC}}\\ \text{or BC = AB = 75 m}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ABD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30° =}\frac{\text{AB}}{\text{DB}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{75}}{\text{DB}}\\ \text{or DB = 75}\sqrt{\text{3}}\\ \text{or DB = BC+CD = 75}\sqrt{\text{3}}\\ \text{or CD = 75}\sqrt{\text{3}}-\text{BC = 75}\sqrt{\text{3}}-\text{75 = 75\hspace{0.17em}(}\sqrt{\text{3}}-\text{1)}\\ \text{Therefore,}\\ \text{distance between the two ships = 75\hspace{0.17em}(}\sqrt{\text{3}}-\text{1) m}\end{array}$

Q.14 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° (see the following figure). Find the distance travelled by the balloon during the interval.


Ans.

$\begin{array}{l}\text{Let the balloon be initially at A and then it moves to B.}\\ \text{Let CD be the girl.}\\ \text{In Δ\hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° =}\frac{\text{AE}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AE}}{\text{CE}}\\ \text{or CE =}\frac{\text{AE}}{\sqrt{\text{3}}}\text{=}\frac{\text{AF}-\text{EF}}{\sqrt{\text{3}}}\text{=}\frac{\text{88.2}-\text{1.2}}{\sqrt{\text{3}}}\text{=}\frac{\text{87}}{\sqrt{\text{3}}}\text{= 29}\sqrt{\text{3}}\\ \text{In Δ\hspace{0.17em}BCG, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30° =}\frac{\text{BG}}{\text{CG}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{88.2}-\text{1.2}}{\text{CG}}\\ \text{or CG = 87}\sqrt{\text{3}}\text{m}\\ \text{Therefore,}\\ \text{distance travelled by the balloon = EG}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= CG-CE}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \hspace{0.33em}= 87}\sqrt{\text{3}}-\text{29}\sqrt{\text{3}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} = 58}\sqrt{\text{3}}\text{m}\end{array}$

Q.15 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 time taken by the car to reach the foot of the tower from this point.

Ans.

$\begin{array}{l}\text{Let AB be a man and BC be a tower. Let the car is initially}\\ \text{at D and after 6 seconds it moves to point E.}\\ \text{In Δ\hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° =}\frac{\text{AC}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AC}}{\text{CE}}\\ \text{or AC =}\sqrt{\text{3}}\text{\hspace{0.17em}CE}...\text{(1)}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ACD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30° =}\frac{\text{AC}}{\text{CD}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{AC}}{\text{CD}}\\ \text{or CD = AC}\sqrt{\text{3}}\\ \text{or CD =}\sqrt{\text{3}}\text{×}\sqrt{\text{3}}\text{\hspace{0.17em}CE = 3CE}\left[\text{from (1), AC =}\sqrt{\text{3}}\text{\hspace{0.17em}CE}\right]\\ \text{or DE+CE = 3CE}\\ \text{or DE = 2CE}\\ \text{or CE =}\frac{\text{DE}}{\text{2}}\\ \text{The car takes 6 seconds to cover the distance DE.}\\ \text{Therefore, it will take 3 seconds to cover the half of the}\\ \text{distance DE i.e., the car will take 3 seconds to reach the}\\ \text{foot of the tower from the point E.}\end{array}$

Q.16 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Ans.

Let AB is a tower. Let C and D respectively are two points at a distance of 4 m and 9 m from the base of the tower.

$\begin{array}{l}\text{Let angles of elevation of the top of the tower from D be θ}\\ \text{and from C be 90°}-\text{θ.}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}(90°}-\text{θ)\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{BC}}\\ \text{or cotθ\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{4}}\\ \text{or AB\hspace{0.33em}=\hspace{0.33em}4cotθ}...\text{(1)}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ABD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}θ\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{BD}}\\ \text{or \hspace{0.17em}tan\hspace{0.17em}θ\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{9}}\\ \text{or \hspace{0.17em}\hspace{0.17em}AB\hspace{0.33em}=\hspace{0.33em}9 tanθ}...\text{(2)}\\ \text{From (1) and (2), we have}\\ \text{9 tanθ\hspace{0.33em}=\hspace{0.33em}4cotθ}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\frac{\text{tanθ}}{\text{cotθ}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{4}}{\text{9}}\\ {\text{or \hspace{0.17em}\hspace{0.17em}tan}}^{\text{2}}\text{θ\hspace{0.33em}=\hspace{0.33em}}\frac{\text{4}}{\text{9}}\\ \text{or \hspace{0.17em}\hspace{0.17em}tanθ\hspace{0.33em}=\hspace{0.33em}}\frac{\text{2}}{\text{3}}\\ \text{On putting this value of tanθ in (2), we get}\\ \text{AB\hspace{0.33em}=\hspace{0.33em}9 tanθ\hspace{0.33em}=\hspace{0.33em}9×}\frac{\text{2}}{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}6 m}\\ \text{Therefore, the height of the tower is 6 m.}\end{array}$