Case study (4 Marks) - MATHS STD 10 Questions - Vidyadip

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Case study (4 Marks)

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Question 14 Marks

The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes 60°. Show that the height of the tower is 129.9 metres. $(\text{Use }\sqrt{3}=1.7323=1.732)$

Answer

Let, height of tower, AB = h
angle of elevation $\angle\text{D}=30^\circ,\ \angle\text{C}=60^\circ$
Distance, DC = 150°
Now, we have to prove that height of tower is 129.9m

In $\triangle\text{ABC},$
$\tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\sqrt{3}}$
In $\triangle\text{ABD},$
$\tan\text{D}=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{AB}}{\text{DC}+\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{150+\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{150+\text{x}}$
$\Rightarrow\ \text{h}=\frac{150+\text{x}}{\sqrt{3}}$
$\Rightarrow\ \text{h}=\frac{150+\frac{\text{h}}{\sqrt{3}}}{\sqrt{3}}$ $[\because\ \sqrt{3}=1.732]$
$\Rightarrow\ \text{h}=129.9$
Hence proved, height of tower is 129.9m

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Question 24 Marks

A man on the deck of a ship is 10m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.

Answer

Height of ship from water level = 10cm = AB
Angle of elevation of top of cliff $\alpha=45^\circ$
Angle of depression of bottom of cliff $\alpha=30^\circ$
Height of cliff CD = 'h' m.
Distance of ship from foot of tower cliff
Height of cliff above ship be 'a' m
Then height of cliff = DX + XC
= (10 + a)m
The above data is represented in form of figure as shown
In right triangle, if one of the included angle is $\theta,$ then
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{CX}}{\text{AX}}$
$\Rightarrow\ 1=\frac{\text{a}}{\text{AX}}$
⇒ AX = 'a' m
$\Rightarrow\ \tan30^\circ=\frac{\text{XD}}{\text{AX}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{10}{\text{AX}}$
$\Rightarrow\ \text{AX}=10\sqrt{3}$
$\therefore\ \text{a}=10\sqrt{3}\text{m}$
Height of cliff $=10+10\sqrt{3}=10(\sqrt{3}+1)\text{m}$
Distance between ship and cliff $=10\sqrt{3}\text{m.}$

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Question 34 Marks

The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10m longer than when it was 60°. Find the height of the tower.

Answer

Let, height of tower, AC = h
angle of elevation $\angle\text{B}=45^\circ,\ \angle\text{D}=60^\circ$
distance BC = 10 + x, CD = x

In $\triangle\text{ABC},$
$\tan\text{B}=\frac{\text{AC}}{\text{BC}}$
$\tan45^\circ=\frac{\text{h}}{10+\text{x}}$
$1=\frac{\text{h}}{10+\text{x}}$
$\text{h}=10+\text{x}$
$\text{x}=\text{h}-10\ .....(\text{i})$
In $\triangle\text{ACD},$
$\tan\text{D}=\frac{\text{h}}{\text{x}}$
$\tan60^\circ=\frac{\text{h}}{\text{x}}$
$\sqrt{3}=\frac{\text{h}}{\text{x}}$
$\text{h}=\sqrt{3}\text{x}\ ......(\text{ii})$
From eq. (i) put the value of x in eq. (ii)
$\text{h}=\sqrt{3}(\text{h}-10)$
$\text{h}=\sqrt{3}\text{h}-10\sqrt{3}$
$\text{h}(\sqrt{3}-1)=10\sqrt{3}$
$\text{h}=\frac{10\sqrt{3}}{\sqrt{3}-1}$
$\text{h}=\frac{10\times1.732}{1.732-1}$
$\text{h}=\frac{17.32}{0.732}$
$\text{h}=23.66$
Hence, height of tower is 23.66m.

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Question 44 Marks

There are two temples, one on each bank of a river, just opposite to each other. One temple is 50m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.

Answer

Let AB and CE are two temples each at the bank of river. The top of the temple CE makes angle of depressions at the top and bottom of tower AB are 30° and 60°
Let CE = 50m and AB = Hm and $\angle\text{CBE}=60^\circ,\ \angle\text{DAE}=30^\circ$
The corresponding figure is as follows,

In $\triangle\text{ADE},$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}\sqrt{3}$
Again in $\triangle\text{BCE},$
$\Rightarrow\ \tan60^\circ=\frac{50}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{50}{\text{x}}$
$\Rightarrow\ 50=\sqrt{3}\times\text{h}\sqrt{3}$
$\Rightarrow\ \text{h}=\frac{50}{3}$
Now the distance between the temples,
$\text{x}=\text{h}\sqrt{3}$
$=\frac{50}{3}\times\sqrt{3}$
$=\frac{50}{\sqrt{3}}$
Therefore, $\text{H}=50-\frac{50}{3}$
⇒ H = 33.33
Hence distance between the temples is $\frac{50}{\sqrt{3}}\text{m}=28.83\text{m}$ and height of temple is 33.33m.

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Question 54 Marks

From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be $\alpha$ and $\beta.$ Show that the height in miles of aeroplane above the road is given by, $\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}.$

Answer

Let A is aeroplane and C and D are two such points that the angles of depression from A are a and p respectively and CD = 1km
Let height of the plane be h

XY || CD
$\therefore\ \angle\text{C}=\alpha,\ \angle\text{D}=\beta$ (Alternate angles)
Let BC = xkm, then BD = (1 - x)km
Now in right $\triangle\text{ACB,}$
$\tan\alpha=\frac{\text{AB}}{\text{BC}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\tan\alpha}$
Similarly in right $\triangle\text{ABD,}$
$\tan\beta=\frac{\text{AB}}{\text{BD}}=\frac{\text{h}}{1-\text{x}}$
$\Rightarrow\ \text{h}=\tan\beta(1-\text{x})$
$\Rightarrow\ \text{h}=\tan\beta-\text{x}\tan\beta$
$=\tan\beta-\frac{\text{h}}{\tan\alpha}\tan\beta$
$\Rightarrow\ \text{h}+\text{h}\frac{\tan\beta}{\tan\alpha}=\tan\beta$
$\Rightarrow\ \text{h}\Big(1+\frac{\tan\beta}{\tan\alpha}\Big)=\tan\beta$
$\Rightarrow\ \text{h}\frac{(\tan\alpha+\tan\beta)}{\tan\alpha}=\tan\beta$
$\Rightarrow\ \text{h}=\frac{\tan\beta+\tan\alpha}{\tan\alpha+\tan\beta}$
$\therefore$ Height of aeroplane $\frac{\tan\beta+\tan\alpha}{\tan\alpha+\tan\beta}\text{km.}$
Hence proved.

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Question 64 Marks

The lower window of a house is at a height of 2m above the ground and its upper window is 4m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.

Answer

Let the window G be 2m above the ground.
Window A be 4m above the window G.
Balloon be at point B above the ground.
$\tan30^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}\sqrt{3}\ ....(\text{i})$
In $\triangle\text{BGD,}$
$\tan60^\circ=\frac{\text{BD}}{\text{GD}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}+4}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}+4}{\sqrt{3}}\ .....(\text{ii})$
From (i) and (ii) we get,
$\text{h}\sqrt{3}=\frac{\text{h}+4}{\sqrt{3}}$
$\Rightarrow\ 3\text{h}=\text{h}+4$
$\Rightarrow\ 2\text{h}=4$
$\Rightarrow\ \text{h}=2$
Hence, the height of the balloon above the ground = 2 + 4 + 2 = 8m.

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Question 74 Marks

From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be $\alpha$ and $\beta.$ If the height of the light house be h metres and the line joining the ships passes through the foot of the light house, show that the distance between the ship is, $\frac{\text{h}(\tan\alpha+\tan\beta)}{\tan\alpha+\tan\beta}\text{meters.}$

Answer

Height of light house = 'h' meters = AB
S and S be two ships on opposite sides of light house $=\alpha$
Angle of depression of 1S from top of light house $=\alpha$
Angle of depression of 2S from top of light house
Required to prove that,
Distance between ships $=\frac{\text{h}(\tan\alpha+\tan\beta)}{\tan\alpha.\tan\beta}\text{meters}$
The above information is represented in the form of figure as shown
In $\triangle\text{ABS}_1$
$\tan\alpha=\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{\text{AB}}{\text{S}_1\text{B}}$
$\text{S}_1\text{B}=\frac{\text{h}}{\tan\alpha}\ ......(1)$
In $\triangle\text{ABS}_2$
$\tan\beta=\frac{\text{AB}}{\text{BS}_2}\Rightarrow\ \text{BS}_2=\frac{\text{h}}{\tan\beta}\ ....(2)$
(1) and (2)
$\Rightarrow\ \text{BS}_1+\text{BS}_2$
$=\frac{\text{h}}{\tan\alpha}+\frac{\text{h}}{\tan\beta}$
$\Rightarrow\ \text{S}_1\text{S}_2=\text{h}\Big\{\frac{1}{\tan\alpha}+\frac{1}{\tan\beta}\Big\}$
$=\frac{\text{h}(\tan\alpha+\tan\beta)}{\tan\alpha.\tan\beta}$
$\therefore$ Distance between ships $=\frac{\text{h}(\tan\alpha+\tan\beta)}{\tan\alpha.\tan\beta}\text{meters}.$

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Question 84 Marks

The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.

Answer

Let aeroplane travelled from A to B in 15 sec
Angle of elevation of points A $\alpha=45^\circ$
Angle of elevation of points B $\beta=30^\circ$
Height of aeroplane from ground = 3000 meters = AP = BQ
Distance travelled in 15 secs = AB = PQ
Velocity (or) speed = distance travelled time the above data is represents is form of figure as shown
In right triangle one of the included angle is $\theta$ then,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan\alpha=\frac{\text{AP}}{\text{XP}}$
$\Rightarrow\ \tan45^\circ=\frac{3000}{\text{XP}}$
$\Rightarrow\ 1=\frac{3000}{\text{XP}}$
$\Rightarrow\ \text{XP}=3000\text{m}$
$\Rightarrow\ \tan\beta=\frac{\text{BQ}}{\text{XQ}}$
$\Rightarrow\ \tan30^\circ=\frac{3000}{\text{XQ}}$
$\Rightarrow\ \text{XQ}=3000\sqrt{3}$
$\Rightarrow\ \text{PQ}=\text{XQ}-\text{XP}$
$=3000(\sqrt{3}-1)\text{m}$
$\Rightarrow\ \text{Speed}=\frac{\text{PQ}}{\text{time}}=\frac{3000(\sqrt{3}-1)}{15}$
$=200(\sqrt{3}-1)$
= 200(0.732)
= 146.4m/sec
Speed of aeroplane = 146.4m/sec.

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Question 94 Marks

The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower.

Answer

Let TR is the tower From a point X, the angle of elevation of T is 60° and 40m above X, from the point Y, the angle of elevation is 45°.

From Y, draw YZ || XR
Let TR = h and XR = YZ = x
The ZR = YX = 40m and TZ = (h - 40)m
Now in right $\triangle\text{TXR},$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{TR}}{\text{XR}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}}{\text{x}}\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\sqrt{3}}\ ....(\text{i})$
Similarly in right $\triangle\text{TYZ,}$
$\tan45^\circ=\frac{\text{TZ}}{\text{YZ}}\Rightarrow\ 1=\frac{\text{h}-40}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}-40\ .......(\text{ii})$
From (i) and (ii)
$\frac{\text{h}}{\sqrt{3}}=\text{h}-40$
$\Rightarrow\ \text{h}=\sqrt{3}\text{h}-40\sqrt{3}$
$\Rightarrow\ \sqrt{3}\text{h}-\text{h}=40\sqrt{3}$
$\Rightarrow\ \text{h}(\sqrt{3}-1)=40\sqrt{3}$
$\Rightarrow\ \text{h}=\frac{40\sqrt{3}}{\sqrt{3}-1}=\frac{40\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$=\frac{40\sqrt{3}(\sqrt{3}+1)}{3-1}=\frac{40\sqrt{3}(\sqrt{3}+1)}{2}$
$=20\sqrt{3}(\sqrt{3}+1)=20\times3+20\sqrt{3}$
= 60 + 20 × 1.732
= 60 + 34.640 = 94.64
Hence height of the tower = 94.64m

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Question 104 Marks

A carpenter makes stools for electricians with a square top of side 0.5m and at a height of 1.5m above the ground. Also, each leg is inclined at an angle of 60° to the ground. Find the length of each leg and also the lengths of two steps to be put at equal distances.

Answer

Let the length of stool, AB = 0.5m, height AC = 1.5m and its leg inclined at an angle of 60° to the ground.
Let length of leg AE = h m

We have to find length of leg, lengths of two steps equal in length.
In $\triangle\text{AEC},\ \angle\text{AEC}=60^\circ$
$\sin60^\circ=\frac{\text{AC}}{\text{AE}}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\frac{1.5}{\text{h}}$
$\Rightarrow\ \text{h}=\frac{3}{\sqrt{3}}$
$\Rightarrow\ \text{h}=1.732$
In $\triangle\text{AGH},\ \angle\text{AGH}=60^\circ$ and AH = 0.5m
$\tan60^\circ=\frac{\text{AH}}{\text{GH}}$
$\Rightarrow\ \sqrt{3}=\frac{0.5}{\text{GH}}$
$\Rightarrow\ \text{GH}=\frac{0.5}{\sqrt{3}}$
$\Rightarrow\ \text{GH}=0.2886$
Total length = 0.5 + (0.2886 × 2) = 1.1077m
In $\triangle\text{APQ},\ \angle\text{APQ}=60^\circ$ and AQ = 1m
$\tan60^\circ=\frac{\text{AQ}}{\text{PQ}}$
$\Rightarrow\ \sqrt{3}=\frac{1}{\text{PQ}}$
$\Rightarrow\ \text{PQ}=\frac{1}{\sqrt{3}}$
$\Rightarrow\ \text{PQ}=0.577$
Total lengths 0.5 + (0.577 × 2) = 1.654m
Hence the length of leg is 1.732m
And lengths of each step are 1.1077m and 1.654m

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Question 114 Marks

The angle of elevation of the top of a tower as observed form a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan32° = 0.6248 and tan63° = 1.9626]

Answer

Let h be height of tower and the angle of elevation as observed from the foot of tower is 32° and observed move towards the tower with distance 100m then angle of elevation becomes 63°.
Let BC = x and CD = 100
Now we have to find height of tower, So we use trigonometrical ratios.

In a triangle ABC,
$\Rightarrow\ \tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan63^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ 1.9626=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{1.9626}$
Again in a triangle ABD,
$\Rightarrow\ \tan\text{D}=\frac{\text{AB}}{\text{BC}+\text{CD}}$
$\Rightarrow\ \tan32^\circ=\frac{\text{h}}{\text{x}+100}$
$\Rightarrow\ 0.6248=\frac{\text{h}}{\text{x}+100}$
$\Rightarrow\ \text{x}+100=\frac{\text{h}}{0.6248}$
$\Rightarrow\ 100=\frac{\text{h}}{0.6248}-\frac{\text{h}}{1.9626}$
$\Rightarrow\ 100=\frac{\text{h}\times1.9626-\text{h}\times0.6248}{0.6248\times1.9626}$
$\Rightarrow\ 100=\frac{\text{h}(1.9626-0.6248)}{0.6248\times1.9626}$
$\Rightarrow\ 100=\frac{\text{h}(1.3378)}{0.6248\times1.9626}$
$\Rightarrow\ 100\times0.6248\times1.9626=\text{h}\times1.3378$
$\Rightarrow\ \text{h}=\frac{100\times0.6248\times1.9626}{1.3378}$
$\Rightarrow\ \text{h}=\frac{122.6232}{1.3378}$
$\Rightarrow\ \text{h}=91.66$
$\Rightarrow\ \text{x}=\frac{91.66}{1.9626}$
$\Rightarrow\ \text{x}=46.7$
So distance of the first position from the tower is = 100 + 46.7 = 146.7m
Hence the height of tower is 91.66m and the desires distance is 146.7m

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Question 124 Marks

An aeroplane is flying at a height of 210m. Flying at this height at some instant the angles of depression of two points in opposite directions on both the banks of the river are 45º and 60º. Find the width of the river. $(\text{Use }\sqrt{3}=1.73)$

Answer

Height of aeroplane AB = 210m

$\tan45^\circ=\frac{\text{AB}}{\text{AQ}}$
$1=\frac{210}{\text{AQ}}$
$\text{AQ}=210\text{m}$
$\text{y}=210\text{m}$
Now, $\tan60^\circ=\frac{\text{AB}}{\text{PA}}$
$\Rightarrow\sqrt{3}=\frac{210}{\text{x}}$
$\text{x}=\frac{210}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=70\sqrt{3}$
Distance below bank = x + y
$=70\sqrt{3}+210$
$121.24+210$
$=331.24\text{m}$

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Question 134 Marks

Two boats approach a light house in mid-sea from opposite directions. The angles of elevation of the top of the light house from two boats are 30° and $45^\circ$ respectively. If the distance between two boats is 100m, find the height of the light house.

Answer

Let $B_1$ be boat 1 and $B_2$ be boat 2 .
Height of light house $=' h$ ' $m=A B$
Distance between $B_1 B_2=100 m$
Angle of elevation of A from $B _1 \alpha=30^{\circ}$
Angle of elevation of B from $B _2 \beta=45^{\circ}$
The above information is represented in the form of figure as shown here,
In $\triangle\text{ABB}_1$
$\tan30^\circ=\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{\text{AB}}{\text{B}_1\text{B}}$
$\text{B}_1\text{B}=\text{AB}\sqrt{3}=\text{h}\sqrt{3}\ .......(1)$
In $\triangle\text{ABB}_2$
$\tan45^\circ=\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{\text{AB}}{\text{B}\text{B}_2}\ ......(2)$
$(1) + (2)$
$\Rightarrow\ \text{B}_1\text{B}+\text{BB}_2=\text{h}\sqrt{3}+\text{h}$
$\Rightarrow\ \text{B}_1\text{B}_2=\text{h}(\sqrt{3}+1)$
$\Rightarrow\ \text{h}=\frac{\text{B}_1\text{B}_2}{\sqrt{3}+1}=\frac{100}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{100(\sqrt{3}-1)}{2}=50(\sqrt{3}-1)$
Height of light house $=50(\sqrt{3}-1)$

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Question 144 Marks

A boy is standing on the ground and flying a kite with 100m of string at an elevation of 30°. Another boy is sanding on the roof of a 10m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.

Answer

Let K be the kite. A and B are two boys flying kites. Boy B is standing on a building 10m high. The string AK of kite of boy A is 100m Let h be the height of the kite and x is the length of string of kite of second boy B.

$\therefore$ KD = (h - 10)m
Now in right $\triangle\text{AKT,}$
$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{KT}}{\text{AK}}$
$\sin30^\circ=\frac{\text{h}}{100}=\frac{1}{2}=\frac{\text{h}}{100}$
$\Rightarrow\ \text{h}=\frac{100}{2}=50\text{m}$
Similarly in right $\triangle\text{KDB,}$
$\sin45^\circ=\frac{\text{KD}}{\text{KB}}\Rightarrow\ \frac{1}{\sqrt{2}}=\frac{50-10}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\frac{40}{\text{x}}\Rightarrow\ \text{x}=40\sqrt{2}$
$=40(1.414)=456560$
$\therefore$ Length of string of second kite $=40\sqrt{2}\text{m}$ or 45.656m

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Question 154 Marks

An aeroplane flying horizontally 1km above the ground is observed t an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.

Answer

An aero plane is flying 1km above the ground making an angle of elevation of aero plane 60°. After 10 seconds angle of elevation is changed to 30°. Let CE = x, BC = y, $\angle\text{AEB}=30^\circ,\ \angle\text{DEC}=60^\circ,$ AB = 1km and CD = 1km. Here we have to find speed of aero plane.
The corresponding figure is as follows,

So we use trigonometric ratios.
In $\triangle\text{DCE},$
$\Rightarrow\ \tan60^\circ=\frac{1}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{1}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{1}{\sqrt{3}}$
Again in $\triangle\text{ABE},$
$\Rightarrow\ \tan30^\circ=\frac{1}{\text{x}+\text{y}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{1}{\text{x}+\text{y}}$
$\Rightarrow\ \text{x}+\text{y}=\sqrt{3}$
$\Rightarrow\ \text{y}=\sqrt{3}-\frac{1}{\sqrt{3}}$
$\Rightarrow\ \text{y}=\frac{2}{\sqrt{3}}$
$\text{Speed}=\frac{\text{distance}}{\text{time}}$
$=\frac{\text{y}}{10\text{ sec}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{10}{60\times60}}$
$=415.68$
Hence the speed of aeroplane is 415.68km/h.

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Question 164 Marks

A moving boat is observed from the top of a 150m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.

Answer

Let the distance BC be xm and CD be ym.

In $\triangle\text{ABC,}$
$\tan60^\circ=\frac{\text{AB}}{\text{BC}}=\frac{150}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{150}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{150}{\sqrt{3}}\ ......(\text{i})$
In $\triangle\text{ABD,}$
$\tan45^\circ=\frac{\text{AB}}{\text{BD}}=\frac{150}{\text{x}+\text{y}}$
$\Rightarrow\ 1=\frac{150}{\text{x}+\text{y}}$
$\Rightarrow\ \text{x}+\text{y}=150$
$\Rightarrow\ \text{y}=150-\text{x}$
Using (i), we get
$\Rightarrow\ \text{y}=150-\frac{150}{\sqrt{3}}=\frac{150(\sqrt{3}-1)}{\sqrt{3}}\text{m}$
Time taken to move from point C to point D is 2 min $=\frac{2}{60}\text{h}=\frac{1}{30}\text{h}$
Now, $\text{Speed}=\frac{\text{Distance}}{\text{Time}}=\frac{\text{y}}{\frac{1}{30}}$
$=\frac{\frac{150(\sqrt{3}-1)}{\sqrt{3}}}{\frac{1}{30}}=1500\sqrt{3}(\sqrt{3}-1)\text{m/h}$
= 1500 × 1.732(1.732 - 1)m/h
= 2598 × .732 = 1902m/h

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Question 174 Marks

The horizontal distance between two poles is 15m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24m, find the height of the first pole. $(\sqrt{3}=1.732)$

Answer

Let AB and CD are two poles and distance between them = 15m, and AB = 24m Angle of elevation from top of pole CD, to pole AB = 30°
From C, draw CE || DB

Let CD = xm, then
CE = DB = 15m and AE = AB - EB
⇒ AE = AB - CD = (24 - x)m
Now in $\triangle\text{ACE}$
$\tan\theta=\frac{\text{AE}}{\text{CE}}$
$\Rightarrow\ \tan30^\circ=\frac{24-\text{x}}{15}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{24-\text{x}}{15}$
$\Rightarrow\ 15=\sqrt{3}(24-\text{x})$
$\Rightarrow\ \frac{15}{\sqrt{3}}=24-\text{x}$
$\Rightarrow\ \text{x}=24-\frac{15}{\sqrt{3}}=24-\frac{15\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$=24-\frac{15\sqrt{3}}{3}=24-5\sqrt{3}$
$=24-5(1.732)=24-8.660$
$=15.34\text{m}$

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Question 184 Marks

The angle of elevation of the top of a chimney from the top of a tower is 60º and the angle of depression of the foot of the chimney from the top of the tower is 30º. If the height of the tower is 40m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question?

Answer

Let AB be the tower and CD be the chimney.
Height of the tower, AB = 40m
Suppose the height of the chimney be h m.
Draw $\text{AE}\bot\text{CD}.$
Here, CE = AB = 40m
DE = CD - CE = (h - 40)m
In right $\triangle\text{AEC},$
$\tan30^\circ=\frac{\text{CE}}{\text{AE}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{40}{\text{AE}}$
$\Rightarrow\ \text{AE}=40\sqrt{3}\text{m}$
In right $\triangle\text{AED},$
$\tan60^\circ=\frac{\text{DE}}{\text{AE}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}-40}{40\sqrt{3}}$
$\Rightarrow\ \text{h}-40=40\sqrt{3}\times\sqrt{3}=120$
$\Rightarrow\ \text{h}=120+40=160\text{m}$
Thus, the height of the chimney is 160m.
Clearly, the height of the chimney meets the pollution norms.
We should follow the pollution control norms and contribute to the cleaniness of the environment.

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Question 194 Marks

Two ships are there in the sea on either side of a light house in such away that the ships and the light house are in the same straight line. The angles of depression of two ships are observed from the top of the light house are 60º and 45º respectively. If the height of the light house is 200m, find the distance between the two ships. $(\text{Use }\sqrt{3}=1.73)$

Answer

Let CD be the light house and A and B be the positions of the two ships.
Height of the light house, CD = 200m
Now,
$\angle\text{CAD}=\angle\text{ADX}=60^\circ$ (Alternate angles)
$\angle\text{CBD}=\angle\text{BDY}=45^\circ$ (Alternate angles)
In right $\triangle\text{ACD},$
$\tan60^\circ=\frac{\text{CD}}{\text{AC}}$
$\Rightarrow\ \sqrt{3}=\frac{200}{\text{AC}}$
$\Rightarrow\ \text{AC}=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}\text{m}$
In right $\triangle\text{BCD},$
$\tan45^\circ=\frac{\text{CD}}{\text{BC}}$
$\Rightarrow\ 1=\frac{200}{\text{BC}}$
$\Rightarrow\ \text{BC}=200\text{m}$
$\therefore$ Distance between the two ships, AB = BC + AC
$=200+\frac{200\sqrt{3}}{3}$
$=200+\frac{200\times1.73}{3}$
$=200+115.33$
$=315.33\text{m}$ (approx)
Hence, the distance between the two ships is approximately 315.33m

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Question 204 Marks

From the top of a tower h meter high, the angles of depression of two objects, which are in the line with the foot of the tower are $\alpha$ and $\beta(\beta>\alpha).$ Find the distance between the two objects.

Answer

Let the distance between two objects is xm
and CD = ym
Given that, $\angle\text{BAX}=\alpha=\angle\text{ABD}$ [alternate angle]
$\angle\text{CAY}=\beta=\angle\text{ACD}$ [alternate angle]
and the height of tower, AD = h m

Now, in $\triangle\text{ACD,}$
$\tan\beta=\frac{\text{AD}}{\text{CD}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{\text{h}}{\tan\beta}\ ....(\text{i})$
and in $\triangle\text{ABD},$
$\tan\alpha=\frac{\text{h}}{\text{x}+\text{y}}\Rightarrow\ \text{x}+\text{y}=\frac{\text{h}}{\tan\alpha}$
$\Rightarrow\ \text{y}=\frac{\text{h}}{\tan\alpha}-\text{x}\ .....(\text{ii})$
From eq. (i) and (ii),
$\frac{\text{h}}{\tan\beta}=\frac{\text{h}}{\tan\alpha}-\text{x}$
$\therefore\ \text{x}=\frac{\text{h}}{\tan\alpha}-\frac{\text{h}}{\tan\beta}$
$=\text{h}\Big(\frac{1}{\tan\alpha}-\frac{1}{\tan\beta}\Big)=\text{h}(\cot\alpha-\cot\beta)$
$\Big[\because\ \cot\theta=\frac{1}{\tan\theta}\Big]$
Which is the required distance between the two objects.

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Question 214 Marks

A window of a house is h metre above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be $\alpha$ and $\beta$ respectively. Prove that the height of the house is $\text{h}(1+\tan\alpha\tan\beta)\text{meters.}$

Answer

Let the height of the other house = OQ = H
and OB = MW = xm
Given that, height of the first house = WB = h = MO
and $\angle\text{QWM}=\alpha,\ \angle\text{OWM}=\beta=\angle\text{WOB}$ [alternate angle]
Now, in $\triangle\text{WOB,}$
$\tan\beta=\frac{\text{WB}}{\text{OB}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\tan\beta}\ ......(\text{i})$
And in $\triangle\text{QWM,}$
$\tan\alpha=\frac{\text{QM}}{\text{WM}}=\frac{\text{OQ}-\text{MO}}{\text{WM}}$
$\Rightarrow\ \tan\text{a}=\frac{\text{H}-\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{H}-\text{h}}{\tan\alpha}\ .....(\text{ii})$
From eq. (i) and (ii),
$\frac{\text{h}}{\tan\beta}=\frac{\text{H}-\text{h}}{\tan\alpha}$
$\Rightarrow\ \text{h}\tan\alpha=(\text{H}-\text{h})\tan\beta$
$\Rightarrow\ \text{h}\tan\alpha=\text{H}\tan\beta-\text{h}\tan\beta$
$\Rightarrow\ \text{H}\tan\beta=\text{h}(\tan\alpha+\tan\beta)$
$\therefore\ \text{H}=\text{h}\Big(\frac{\tan\alpha+\tan\beta}{\tan\beta}\Big)$
$\text{H}=\text{h}\Big(1+\tan\alpha.\frac{1}{\tan\beta}\Big)$
$=\text{h}(1+\tan\alpha.\cot\beta)$ $\Big(\because\ \cot\theta=\frac{1}{\tan\theta}\Big)$
Hence, the required height of the other house is $\text{h}(1+\tan\alpha.\cot\beta).$
Hence proved.

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Question 224 Marks

An observer, 1.5m tall, is 28.5m away from a a tower 30m high. Determine the angle elevation of the top of the tower from his eye.

Answer

Height of observer = AB = 1.5m
Height of tower = PQ = 30m
Height of tower above the observe eye = 30 - 1.5
QX = 28.5m
Distance between tower and observe XB = 28.5m
$\theta$ be angle of elevation of tower top from eye
The above data is represented in form of figure as shown from figure,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan\theta=\frac{\text{QX}}{\text{BX}}=\frac{28.5}{28.5}=1$
$\Rightarrow\ \theta=\tan^{-1}(1)=45^\circ$
Angle of elevation = 45°

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Question 234 Marks

The angles of depression of two ships from the top of a light house and on the same side of it are found to be $45^\circ$ and $30^\circ$ respectively. If the ships are 200m apart, find the height of the light house.

Answer

Height of light house AB = 'h' meters
Let $S_1$ and $S_2$ be ships distance between ships $S_1S_2$
Angle of depression of $S_1$ $[\alpha=30^\circ]$
Angle of depression of $S_2$ $[\beta=45^\circ]$
The above data is represented in form of figure as shown
In $\triangle\text{ABS}_2$
$\tan\beta=\frac{\text{AB}}{\text{BS}_2}$
$\tan45^\circ=\frac{\text{h}}{\text{BS}_2}$
$\text{BS}_2=\text{h}\ ......(1)$
In $\triangle\text{ABS}_1$
$\tan\alpha=\frac{\text{AB}}{\text{BS}_1}$
$\tan30^\circ=\frac{\text{h}}{\text{BS}_1}$
$\text{BS}_1=\text{h}\sqrt{3}\ .....(2)$
(2) and (1)
$\Rightarrow\ \text{BS}_1-\text{BS}_2=\text{h}(\sqrt{3}-1)$
$\Rightarrow\ 200=\text{h}(\sqrt{3}-1)$
$\Rightarrow\ \text{h}=\frac{200}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\frac{200}{2}(\sqrt{3}+1)=100(\sqrt{3}+1)\text{ meters}$
$\text{h}=100(1.732+1)=273.2\text{ meters}$
Height of light house = 273.2 meters

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Question 244 Marks

A straight highway leads to the foot of a tower of height 50m. From the top of the tower, the angles of depression of two cars standing on the highway are 30° and 60° respectively. What is the distance the two cars and how far is each car from the tower?

Answer

Let TR be the tower and A and B are two cars on the road making angles of elevation with T the top of tower as 30° and 60°.
Height of the tower TR = 50m
Let AR = x and BR = y
Now in right $\triangle\text{TAR},$

$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{TR}}{\text{AR}}$
$\tan30^\circ=\frac{50}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{50}{\text{x}}$
$\Rightarrow\ \text{x}=50\sqrt{3}\ ......(1)$
Similarly in right $\triangle\text{TRB,}$
$\tan60^\circ=\frac{\text{TR}}{\text{BR}}=\frac{50}{\text{y}}$
$\Rightarrow\ \sqrt{3}=\frac{50}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{50}{\sqrt{3}}\ .....(2)$

AB = AR - BR = x - y

$=\Big(50\sqrt{3}-\frac{50}{\sqrt{3}}\Big)=50\Big(\sqrt{3}-\frac{1}{\sqrt{3}}\Big)$
$=50\Big(\frac{3-1}{\sqrt{3}}\Big)=\frac{50\times2}{\sqrt{3}}$
$=\frac{100\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{100\sqrt{3}}{3}$
$=\frac{100(1.732)}{3}=\frac{173.2}{3}=57.7\text{m}$

Distance of A car $=\text{x}=50\sqrt{3}=50(1.732)$

= 86.600 = 86.65m
Distance of B car $=\text{y}=\frac{50}{\sqrt{3}}=\frac{50\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$=\frac{50\sqrt{3}}{3}=\frac{86.6}{3}$
= 28.83m

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Question 254 Marks

A man standing on the deck of a ship, which is 8m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.

Answer

Let M is the man on the deck MN such that MN = 8m. AB is the hill From M, the angle of elevation of A is 60° and angle of depression of B is 30°

Draw MC || NB
$\therefore\ \angle\text{MBN}=\angle\text{CMB}$ (alternate angles)
Let AB = h, NB = MC = x, AC = h - 8 $(\because$ CB = MN = 8m$)$
Now in right $\triangle\text{AMC,}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{AC}}{\text{MC}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}-8}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}-8}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}-8}{\sqrt{3}}\ ......(\text{i})$
Similarly in right $\triangle\text{MBN,}$
$\tan30^\circ=\frac{\text{MN}}{\text{NB}}=\frac{8}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{8}{\text{x}}\Rightarrow\ \text{x}=8\sqrt{3}\ .....(\text{ii})$
From (i) and (ii)
$\Rightarrow\ 8\sqrt{3}=\frac{\text{h}-8}{\sqrt{3}}$
⇒ 8 × 3 = h - 8
⇒ 24 = h - 8 ⇒ h = 24 + 8 = 32
and $\text{x}=8\sqrt{3}$ or 8(1.732) = 13.858m
$\therefore$ Height of the hill = 32m
and distance between the ships = 13.858m

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Question 264 Marks

A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angles of depression of the foot of the tower at a point b metres just above A is $\beta.$ Prove that the height of the tower is $\text{b}\tan\alpha\cot\beta.$

Answer

Let TR is the tower which subtends angle $\alpha$ at a point A on the same plane AB = b and angle of depression of R from B is $\beta$

$\because$ BX || AR
$\therefore\ \angle\text{ARB}=\angle\text{XBR}=\beta$
Let height of tower TR = h
and AR = x
Now in right $\triangle\text{ATR,}$
$\tan\alpha=\frac{\text{h}}{\text{x}}\Rightarrow\ \text{x}=\frac{\text{h}}{\tan\alpha}\ ....(\text{i})$
Similarly in right $\triangle\text{ABR,}$
$\tan\beta=\frac{\text{AB}}{\text{AR}}=\frac{\text{b}}{\text{x}}\Rightarrow\ \text{x}=\frac{\text{b}}{\tan\beta}\ ....(\text{ii})$
From (i) and (ii)
$\frac{\text{h}}{\tan\alpha}=\frac{\text{b}}{\tan\beta}$
$\Rightarrow\ \text{h}=\text{b}\frac{\tan\alpha}{\tan\beta}=\text{b}\tan\alpha\cot\beta$
Hence height of the tower $=\text{b}\tan\alpha\cot\beta$
Hence proved.

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Generate Paper Free All Some Applications of Trigonometry topics

Case study (4 Marks) - MATHS STD 10 Questions - Vidyadip