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 °.
It can be observed from the figure that AB is the pole.\\ $ In \Delta{ABC}, \\ \frac{AB}{ AC} = \sin30^o \\ \frac{AB}{20} = \frac{1}{2} \\ AB= \frac{20}{2} = 10 \\ Therefore the height of the pole is 10 m.$
2 A tree breaks due to storm and 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.
Let AC was a original tree. Due to storm, it was broken into the parts.The broken parts A'B is making $30^o$ with the ground. $ In \Delta A'BC, \\ \dfrac{BC}{A'C} = \tan30^o \\ \dfrac{BC}{8} = \dfrac{ 8}{ \sqrt{3}} \\ BC=\left(\dfrac{8}{\sqrt{3}}\right)m \\ \dfrac{A'C}{A'B} = \cos 30^o \\ \dfrac{8}{A'B} = \dfrac{\sqrt{3}}{2} \\ A'B = \left(\dfrac{16}{\sqrt{3}}\right)m \\ Height of tree = A'B +BC \\ = \left(\dfrac{16}{\sqrt{3}} + \dfrac{8}{\sqrt{3}}\right) = \dfrac{24}{\sqrt{3}}m \\ = 8\sqrt{3}m $
3 A contractor plans to install two slides for the children to play i a park. For the children below the age of $5$ years, she prefers to have a slide whose top is at a height of $1.5m$, and is inclined at an angle of $30^o$ to the ground, whereas for the elder children she wants to have a steep slide at a height of $3m$, and inclind at an angle of $60^o$ to the ground. what should be the length of the slide in each case?
It can be observed that AC and PR are the slides for younger and elder children respectively. $ In \Delta ABC, \\ \\ \dfrac{AB}{AC} = \sin30^o \\ \dfrac{1.5}{AC} = \dfrac{1}{2} \\ AC = 3m \\ \ \ \ \ \ \ \ \ \ \\$
$In \Delta PQR, \\ \dfrac{PQ}{PR} = \sin60^o \\ \dfrac{3}{PR} = \dfrac{\sqrt{3}}{2} \\ PR = \dfrac{6}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}m \\ Therefore, the lengths of these slides are 3m and 2\sqrt{3}m.$
4 The angle of elevation of the top of a tower from a point on the ground, which is $30m$ away from the foot of the tower is $30^o$.Find the height of the tower.
Let AB be the tower and the angle of elevation from point C(on ground) is $30^o$. $ In 'ABC, \\ \dfrac{AB}{BC} = \tan30^o \\ \dfrac{AB}{30} = \dfrac{1}{\sqrt{3}} \\ AB = \dfrac{30}{\sqrt{3}} = 10{\sqrt{3}}m \\ Therefore the height of the tower is 10\sqrt{3}m.$
5 A kite is flying at a height of $60m$ abov 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.
Let $k$ be the kite and the string is tied to point P on the ground. $ In \Delta KLP, \\ \dfrac{KL}{KP} = \sin60^o \\ \dfrac{60}{KP} = \dfrac{\sqrt{3}}{2} \\ KP = \dfrac{120}{\sqrt{3}} = 40\sqrt{3}m \\ Hence,the length of the string is 40\sqrt{3}m.$
6 A $1.5m$ 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 $30^o$ to $60^o$ as he walks towards the building. Find the distance he walked towards the building.
Let the boy was standing at point S initially.He walked towards the building and reached at point $T$. It can be observed that $ PR = PQ-RQ \\ = \left({30-1.5}\right)m = 28.5m = \dfrac{57}{2}m \\ \ \ \ \ \\ In \Delta PAR, \\ \dfrac{PR}{AR} = \tan30^o \\ \dfrac{57}{2AR} = \dfrac{1}{\sqrt{3}} \\ AR = \left({\dfrac{57*\sqrt{3}}{2}}\right)m \\ In \Delta PRB, \\ \dfrac{PR}{BR} = \tan60^o \\ \dfrac{57}{2BR} = \sqrt{3} \\ BR = \dfrac{57}{2 \sqrt{3}} = \left( \dfrac{19 \sqrt{3}}{2} \right)m \\ \ \ \ \ \ \ \\ ST = AB \\ = AR-BR = \left( \dfrac{57 \sqrt{3}}{2} - \dfrac{19 \sqrt{3}}{2}m \right) = \left( \dfrac{38 \sqrt{3}}{2} \right)m = 19 \sqrt{3}m $ Hence, he walked $19\sqrt{3}m$ towards the building.
7 From a point on the ground, the angles of the bottom and the top of a transmission tower fixed at the top of a $20 m$ high building are $45^o$ and $60^o$ respectively. Find the height of the tower.
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 \Delta BCD, \\ \dfrac{BC}{CD} = \tan45^o \dfrac{20}{CD} = 1 \\ CD = 20m \\ \ \ \ \ \ In \Delta ACD, \\ \dfrac{AC}{CD} = \tan60^o \\ \dfrac{AB+BC}{CD} = \sqrt{3} \\ \dfrac{AB+20}{CD} = \sqrt{3} \\ AB=\left(20\sqrt{3} - 20\right)m\\ = 20\left(\sqrt{3} -1\right)m \\$ Therefore, the height of the transmission tower is $20\left(\sqrt{3} - 1 \right)m$.
8 A statue $1.6m$ tall, stands on a top of pedestal,from a point on the ground, the angle of elevation of the top of statue is $60^o$ and from the same point the angle of elevation of the top of the pedestal is $45^o$ . Find the height of the pedestal.
Let AB the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured. $ In \Delta BCD, \\ \dfrac{BC}{CD} = \tan45^o \\ \dfrac{BC}{CD} = 1\\ BC=CD\\ \ \ \ \ \ \\ In \Delta ACD, \\ \dfrac{AB+BC}{CD} = \tan60^o \\ \dfrac{AB+BC}{CD} = \sqrt{3} \\ 1.6 +BC =BC\sqrt{3} \ \ \ \ \left[As CD = BC \right] \\ BC\left(\sqrt{3} - 1\right) = 1.6 \\ BC = \dfrac{\left(1.6\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \ \ \ \ \left[By Rationalization\right] \\ = \dfrac{1.6\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2 - \left(1\right)^2} \\ \dfrac{1.6\left(\sqrt{3}+1\right)}{2} = 0.8\left(\sqrt{3}+1\right)\\ $ Therefore, the height of the pedestal is $0.8\left(\sqrt{3}+1\right)m$
9 The angle of the elevation of the top of a building from the foot of the tower is $30^o$ and the angle of the elevation of the top of the tower from the foot of the building is $60^o$. If the tower is $50 m$ high,find the height of the building.
Let AB be the building CD be the tower. $ In\Delta CDB, \\ \dfrac{CD}{BD} = \tan60^o \\ \dfrac{50}{BD}= \sqrt{3} \\ BD= \dfrac{50}{\sqrt{3}} \\ \ \ \ \ \ \\ In \Delta ABD, \\ \dfrac{AB}{BD} = \tan30^o \\ AB= \dfrac{50}{\sqrt{3}}*\dfrac{1}{\sqrt{3}} = \dfrac{50}{3} = 16\dfrac{2}{3} \\$ Threrefore, the height of the building is $16\dfrac{2}{3}m$.
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 $60^o$ and $30^o$,respectively. Find the height of the poles and the distance of the point from the poles.
Let AB and CD be the poles and $O$ is the point from where the elevation angles are measured. $ In \Delta CDO, \\ \dfrac{AB}{BO} = \tan60^o \\ \dfrac{AB}{BO} = \sqrt{3} \\ BO = \dfrac{AB}{\sqrt{3}} \\ \\ In \Delta CDO, \\ \dfrac{CD}{DO} = \tan30^o \\ \dfrac{CD}{80-BO} = \dfrac{1}{\sqrt{3}} \\ CD\sqrt{3} = 80-BO \\ Cd\sqrt{3} = 80-\dfrac{AB}{\sqrt{3}} \\ CD\sqrt{3}+\dfrac{AB}{\sqrt{3}} = 80 \\ Since the poles are of equal heights. \\ CD=AB \\ CD\left[\sqrt{3}+\dfrac{1}{\sqrt{3}}\right] = 80\\ CD\left(\dfrac{3+1}{\sqrt{3}}\right) = 80 \\ CD = 20\sqrt{3}m\\ BO = \dfrac{AB}{\sqrt{3}} = \dfrac{CD}{\sqrt{3}} = \left(\dfrac{20\sqrt{3}}{\sqrt{}3}\right)m = 20 m\\ DO = BD-BO = \left(80-20\right)m = 60 m $ Therefore,the height of the poles is $20\sqrt{3}$ and the point is $ 20 m $ and $60 m$ far from these poles.
11 A TV tower stands vertically on a bank of a canal.From a point on the other bank directly opposite the lower the angle of elevation of the top of the tower is $60^o$ . 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 $30^o$ . Find the height of the tower and the width of the canel.
$ In \Delta ABC,\\ \dfrac{AB}{BC} = \tan60^o \\ \dfrac{AB}{BC} = \sqrt{3} \\ BC = \dfrac{BC}{\sqrt{}3} \\ \ \ \ \ \\ In \Delta ABD , \\ \dfrac{AB}{BD} = \tan30^o \\ \dfrac{Ab}{BC+CD} = \dfrac{1}{\sqrt{3}} \\ \dfrac{AB}{\dfrac{AB}{\sqrt{3}}+20} = \dfrac{1}{\sqrt{3}} \\ \dfrac{AB\sqrt{3}}{AB+20\sqrt{3}} = \dfrac{1}{\sqrt{3}} \\ 3AB = AB+20\sqrt{3} \\ 2AB= 20\sqrt{3} \\ AB = 10\sqrt{3}m \\ BC = \dfrac{AB}{\sqrt{3}} = \left(dfrac{10\sqrt{3}}{sqrt{3}}\right)m = 10 m $ Therefore, the height of the tower is $10\sqrt{3} m $ and the width of the canal is $10 m $ .
12 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.
Let $AB$ be a building and Cd be a cable tower. $ In \Delta ABD, \\ \dfrac{AB}{BD} = \tan45^o \\ \dfrac{7}{BD} = 1 \implies BD = 7m\\ \\ In \Delta ACE, \\ AE = BD = 7m \\ \\ \dfrac{CE}{AE} = \tan60^o \\ \dfrac{CE}{7 } = \sqrt{3} \\ CE = 7\sqrt{3}m \\ CD = CE +ED = \left(7\sqrt{3}+7\right) m \\ \ \ \ \ \ \ \ = 7\left(\sqrt{3}+1\right) m \\ $ Therefore, the height of the cable tower is $\left(7\sqrt{3}+7\right) m.
13 As observed from the top of a $75 m $ high lighthouse from the sea-level, the angles of depression of two ships are $30^o$ and $45^o$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Let $AB$ be the lighthouse and the two ships be at point C and D respectively. $ In \Delta ABC, \\ \dfrac{AB}{BC} = \tan45^o \\ \dfrac{75}{BC} = 1 \\ BC = 75 m \\ \\ In \Delta ABD, \\ \dfrac{AB}{BD} = \tan30^o \\ \dfrac{75}{BC+CD} = \dfrac{1}{\sqrt{3}} \\ \dfrac{75}{75+CD} = \dfrac{1}{\sqrt{3}} \\ 75\sqrt{3} = 75+CD \\ 75\left(\sqrt{3}-1\right)m = CD \\$ Therefore, the distance between the two ships is $75\left(\sqrt{3}=1\right) m $.
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 $60^o$ . After some time, the angle of elevation reduces to $30^o$. Find the distance travelled by the balloon during the interval.
Let the initial position of $A$ of balloon change to $B$ after some time $CD$ be the girl. $ In \Delta ACE, \\ \dfrac{AE}{CE} = \tan60^o \\ \dfrac{88.2-1.2}{CE} = \sqrt{3} \\ \dfrac{87}{CE} = \sqrt{3} \\ CE = \dfrac{87}{\sqrt{3}} = 29\sqrt{3}m \\ \ \ \ \ \ \\ In \Delta BCG, \\ \dfrac{BG}{CG} = \tan30^o \\ \dfrac{BH-GH}{CG} = \dfrac{1}{\sqrt{3}} \\ \dfrac{88.2 - 1.2}{CG} = \dfrac{1}{\sqrt{3}} \\ 87\sqrt{3}m = CG\\ Distance travelled by balloon = EG = CG- CE \\ = \left(87\sqrt{3} - 29\sqrt{3}\right ) M \\ = 58\sqrt{3}m. $
15 A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as 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.
Let AB be the tower. Initial position of the car is C, which changes to D after six seconds. $In \Delta ADB , \\ \dfrac{AB}{DB} = \tan60^o \\ \dfrac{AB}{DB} = 3 \\ DB = \dfrac{AB}{\sqrt{3}} \\ \\ In \Delta ABC, \\ \dfrac{AB}{BC} = \tan30^o \\ dfrac{AB}{BD+DC} = \dfrac{1}{\sqrt{3}} \\ AB 3 = BD + DC \\ AB\sqrt{3} = \dfrac{AB}{\sqrt{3}} + DC \\ DC= AB\sqrt{3}-\dfrac{AB}{\sqrt{3}}= Ab\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right) \\ = \dfrac{2AB}{\sqrt{3}} \\$ Time taken by the car to travel distance DC $\left(i.e. \dfrac{2AB}{\sqrt{3}}\right) = 6 seconds $ \\ Time taken by the car to travel distance DB $\left(i.e.,\dfrac{AB}{\sqrt{3}}\right) = \dfrac{6}{\dfrac{2AB}{\sqrt{3}}}*\dfrac{AB}{\sqrt{3}} = \dfrac{6}{2} = 3 seconds. $.
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.
17 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 $6 m$. $\\$
Let $A Q$ be the tower and R, S are the points $4 m$, $9 m$ away from the base of the tower respectively.
$ In \Delta AQR, $
The angles are complementary.
Let $A Q$ be the tower and R, S are the points $4 m$, $9 m$ away from the base of the tower respectively. $\\$ The angles are complementary. Therefore, if one angle is$\theta,$ the other will be $90 - \theta.$ $\\$ $ In \Delta AQR, $ $\\$ $\dfrac{A Q}{QR} = \tan \theta $ $\\$ $\dfrac{A Q}{4} = \tan \theta \quad \quad ..............(i)$ $\\$ $ In \Delta AQS, $ $\\$ $\dfrac{A Q}{SQ} = \tan (90 - \theta) $ $\\$ $\dfrac{A Q}{9} = \cot \theta \quad \quad ..............(ii)$ $\\$ On multiplying equations $(i)$ and $(ii)$, we obtain $\\$ $\left(\dfrac{AQ}{4}\right) \left(\dfrac{AQ}{9}\right) = (\tan \theta) (\cot \theta) $ $\\$ $ \dfrac{AQ^2}{36} = 1 $ $\\$ $ AQ^2 = 36 $ $\\$ $ AQ = \sqrt {36} = \pm 6 $ $\\$ However, height cannot be negative. $\\$ Therefore, the height of the tower is $ 6 m.$ $\\$