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Chapter 9: Some Applications of Trigonometry

1. If sun’s elevation is 45°. Find the length of the shadow of a pole of height 8 m.

In rt. Δ PQR, \(\angle B=90°\) and \(\angle QPR= 45°\)

1

∴ tan 45° = \(\frac{QR}{PQ}\)

1 = \(\frac{8}{PQ}\)

\(\Rightarrow AB=\frac{8}{1}\) = 8m

 

2. A pole 8 metre high cast a shadow 8 m on the ground. Find the sun’s elevation.

In rt. Δ PQR, \(\angle B=90°\)

2

∴ PQ = 8 m

QR = 8 m

Let \(\angle QPR=\theta\)

∴ tan \(\theta\) = \(\frac{QR}{PQ}\)

= \(\frac{QR}{PQ}\) = 1

\(\theta\) = 45°

 

3. If the angle of elevation of the top of a hill from two points at distances p and q from the base and in the same straight line with it are complementary. Find the height of the hill.

Let   PQ = h be the height of the hill,

QR = b, QS = a, such that \(\angle PSQ = \theta\)

\(\Rightarrow \angle PRQ= 90°-\theta\)

3

In rt. Δ PQS, we have

\(\frac{h}{a}=tan\) \(\theta\)

h = a tan \(\theta\)          …..(1)

In rt. Δ PQR, we have

\(\frac{h}{b}=tan(90°-\theta )\) = cot \(\theta\)

h = b cot \(\theta\)         …..(2)

Multiplying (1) and (2), we get

\(h^{2}\) = ab tan \(\theta\) cot \(\theta\) = ab

\(\Rightarrow h = \sqrt{ab}\)

 

4. From the figure, find the angle of depression of point Z from the point R.

4

\(\angle YRS=\angle XYR\) (alternative.int. \(\angle a\) )

\(\Rightarrow \angle YRS\) = \(60°\)

\(\Rightarrow \angle ZRS+\angle YRZ=60°\) \(\Rightarrow \angle ZRS+30°=60°\) \(\Rightarrow \angle ZRS=30°\)

 

5. Find the angle of elevation of the moon when the shadow of a tree h metres high is \(\sqrt{3}\)h metres long.

5

The height of the pole be h m and length of its shadow is \(\sqrt{3}\)h

Let \(\theta\) be the elevation of the sun.

Consider rt.agl Δ ABC

tan \(\theta\) = \(\frac{h}{\sqrt{3}h}= \frac{1}{\sqrt{3}}\)

= tan \(30°\)

\(\Rightarrow\)   \(\theta\) = \(30°\)

 

 6. A ladder 14 metres long just reaches the top of a wall. If the ladder makes an angle of 45° with the wall, find the height of the wall.

Here, length of the ladder is 14m and angle of elevation is 90°-45°=45°.

Let h m be the height of the wall

6

Consider rt.agl Δ PQR

\(\frac{QR}{PR}\) = sin 30°

\(\frac{h}{14}=\frac{\sqrt{2}}{2}\)

h = \(\frac{14\sqrt{2}}{2}\) m

∴ The length of the wall is \(7\sqrt{2}\)

 

 7. An observer 1.8 m tall is 20.8 m away from a tower 22.6 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

7

Here, PQ be the observer and RS be the tower of height 24 m.

PQ = TS = 1.8 m

RT = RS – TS

= 22.6 – 1.8

= 20.8 m

Now, Consider rt.agl ΔPTR, we have

\(\frac{RT}{AT}\) = tan \(\theta\)

\(\frac{20.8}{20.8}\) = tan \(\theta\)

1 = tan \(\theta\)

tan \(45°\) = tan \(\theta\)

\(\theta\) =45°

 

8.From a balloon vertically above a straight road, the angles of depression of two cars at an instant is found to be 45° and 60°. If the car are 100 m apart, find the height of the balloon.

Let the height of the balloon at P be h metres. Let A and B be the two cars. Thus, AB = 100m.

8

∠ PAQ= 45° and ∠ PBQ=60°

Now, Consider rt.agl ΔAQP,

\(\frac{PQ}{AQ}\) = tan \(45°\)

\(\Rightarrow\)         \(\frac{PQ}{AQ}\) = 1

\(\Rightarrow\)         PQ = AQ= h m

Now, Consider rt.agl ΔPBQ,

\(\frac{PQ}{BQ}\) = tan \(60°\)

\(\frac{PQ}{BQ}\) = \(\sqrt{3}\)

\(\frac{h}{h-100}\) = \(\sqrt{3}\)

h = \(\sqrt{3}h-\sqrt{3}(100)\)

\((\sqrt{3}-1)h\) = \(\sqrt{3}(100)\)

h = \(\frac{\sqrt{3}(100)}{\sqrt{3}-1}\)

h = \(\frac{\sqrt{3}(100)}{\sqrt{3}-1}\) \(\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

h = \(\frac{\sqrt{3}(100)\left ( \sqrt{3}+1 \right )}{2}\)

= \(50(3+\sqrt{3})\)

 

 9. The shadow of a tower on a level surface is 40 m long when moon’s elevation is 30° then when it is 60°.Find the height of the tower.

XY=40 m, \(\angle RXS\) =30°, \(\angle RYS\) = 60°

9

Now, Consider rt.agl ΔRYS

\(\frac{RY}{SY}\) = tan 60°

\(\Rightarrow\)         \(\frac{RY}{SY}\) = \(\sqrt{3}\)

\(\Rightarrow\)                                   h = \(\sqrt{3}\) SY

Consider rt.agl ΔRXQ

\(\frac{RX}{SX}\) = tan 30°

\(\Rightarrow\)    \(\frac{RX}{XY+YS}\) = \(\frac{1}{\sqrt{3}}\)

\(\Rightarrow\)    \(\frac{h}{40+YS}\)     = \(\frac{1}{\sqrt{3}}\)

\(\Rightarrow\)    \(\sqrt{3}\)h = 40 + YS

\(\Rightarrow\)    \(\sqrt{3}\)h = 40 + \(\frac{h}{\sqrt{3}}\)

3h = 40\(\sqrt{3}\) + h

2h = 40\(\sqrt{3}\)

h = 20\(\sqrt{3}\)

∴ The height of the tower is 20\(\sqrt{3}\) m

 

10. From the top of a tree h m high, the angle of depression of two objects, which are in the line with foot of the tree are \(\gamma\) and \(\delta\) (\(\delta\)> \(\gamma\)). Find the distance between the two objects.

Let RS be the tree of height h m, X and Y are two objects such that

\(\angle RXS\) = \(\gamma\) , \(\angle RYS\) = \(\delta\)

10

Now, Consider rt.agl ΔYSR

\(\frac{RS}{YS}\) = tan \(\delta\)

\(\Rightarrow\)    RS = YS tan \(\delta\)     …..(i)

Now, Consider rt.agl ΔXSR

\(\frac{RS}{XS}\) = tan \(\gamma\)

\(\Rightarrow\) \(\frac{RS}{XY+YS}\) = tan \(\gamma\)

\(\Rightarrow\)         RS = (XY+YS) tan \(\gamma\)     …..(ii)

\(\Rightarrow\) \(\frac{h}{tan \gamma }\) = XY + \(\frac{RS}{tan \delta }\)     [using (i)]

\(\Rightarrow\) h cot \(\gamma\) = XY + h cot \(\delta\)

\(\Rightarrow\)

XY = h (cot \(\gamma\) + cot \(\delta\))

 

11. The angle of elevation of the top of a tower from two distinct points x and y from its foot are complementary. Prove that the height of the tower is \(\sqrt{xy}\).

Let h m be the height of the tower PQ. X and Y are two distinct points from its foot, such that PX = x and PY = y

11

Now, Consider rt.agl ΔXQP, we obtain

\(\frac{PQ}{PX}\) = tan \(\theta\)

\(\frac{h}{x}\) = tan \(\theta\)

h = x tan \(\theta\)     …..(i)

Consider rt.agl ΔYQP, we obtain

\(\frac{PQ}{PY}\) = tan \((90°-\theta )\)

\(\frac{h}{y}\) = cot \(\theta\)

h = y \(\times \frac{1}{tan\theta }\)   …..(ii)

Multiply (I) and (ii), we get

\(h^{2}\) = x tan \(\theta\) \(\times y\times \frac{1}{tan\theta }\) = xy

h = \(\sqrt{xy}\)

∴ The height of the tower is \(\sqrt{xy}\).

 

12. The angle of elevation of the top of a tower from certain point at 30°. If the observer moves 10 m towards the tower, the angle of elevation of the top increases by 15° . Find the height of the tower.

Let h m be the height of the tower RS and P, Q are two points 10 m apaet, such that

\(\angle SPR= 30° and \angle SQR=45°\).

12

Now, Consider rt.agl ΔQRS

\(\frac{SR}{QR}\) = tan 45°

\(\Rightarrow\)                  SR = QR = h m

Now, Consider rt.agl ΔPRS

\(\frac{SR}{PR}\) = tan 30°

\(\Rightarrow\) \(\frac{h}{10+h}\) = \(\frac{1}{\sqrt{3}}\)

\(\Rightarrow\)         \(\sqrt{3}\) h = 10 + h

(\(\sqrt{3}-1\)) h = 10

\(\Rightarrow\) h = \(\frac{10}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\) = \(\frac{10(\sqrt{3}+1)}{2}\)

∴ The height of the tower is 5(\(\sqrt{3}\)+1)

 

13. The angle of elevation of the top of a vertical tower from a point on the groung is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Let XY be the vertical tower of height h m.

C and D are two points 10 m apart such that,

13

CD = ZY = 10 m \(\angle XCZ= 45° and \angle XBY=60°\)

XZ = XY – ZY = (h – 10) m

Consider rt.agl ΔDYX

\(\frac{XY}{DY}\) = tan 60°

\(\frac{h}{DY}\) = \(\sqrt{3}\)

\(\Rightarrow\) DY = \(\frac{h}{\sqrt{3}}\)

Consider rt.agl ΔCZX

\(\frac{XZ}{CZ}\) = tan 45°

XZ = CZ

H – 10 = \(\frac{h}{\sqrt{3}}\)

h \(\left ( 1-\frac{1}{\sqrt{3}} \right )\) = 10

h\(\left ( \frac{\sqrt{3}-1}{\sqrt{3}} \right )\) = 10

h = \(\frac{10\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

= \(\frac{10\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

= 5(3 + \(\sqrt{3}\))m

 

14. The angle of elevation of the top of a tower 27 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and the height of the other tower.

Let us assume that PQ and RS be the two tower of height  h m and 27 m respectively.

\(\angle RQS\) = 60° \(\angle PSQ\) = 30°

14

Consider rt.agl ΔRQS

\(\frac{RS}{QS}\) = tan 60°

\(\frac{27}{QS}\) = \(\sqrt{3}\)

\(\Rightarrow\) QS = \(\frac{27}{\sqrt{3}}\)     …..(i)

Consider rt.agl ΔPSQ

\(\frac{PQ}{QS}\) = tan30°

\(\Rightarrow\) \(\frac{h}{QS}\) = \(\frac{1}{\sqrt{3}}\)

\(\Rightarrow\)   \(\sqrt{3}\) h = QS     …..(ii)

From (i) and (ii), we have

\(\sqrt{3}\) h = \(\frac{27}{\sqrt{3}}\)

\(\Rightarrow\) h = \(\frac{27}{3}\) = 9 m

From (ii), we get QS = 9\(\sqrt{3}\) m

∴ The distance between two towers is 9\(\sqrt{3}\) m and the height of  the other tower is 9 m.

 

15. The lower window of a house is at a height of 3 m above the ground and its upper window is 6 m 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 from the ground.

Let P and Q be the positions of two windows of a house, such that PR = 3 m, QR = 9 m. Let U be the position of balloon of height h m above the ground. \(\angle UPS\) = 60°, \(\angle UQT\) = 30°, US = h – 3 and UT = h – 9

15

Consider rt.agl ΔQTU

\(\frac{UT}{QT}\) = tan 30º

h – 9 = QT \(\times \frac{1}{\sqrt{3}}\)

\(\Rightarrow\) \(\sqrt{3}\) (h – 8) = QT      …..(i)

Consider rt.agl ΔUPS

\(\frac{US}{PS}\) = tan 60º

h – 3 = PS \(\times \sqrt{3}\)

\(\Rightarrow\)  \(\frac{h-3}{\sqrt{3}}\) = PS      …..(ii)

Also, QT = PS     …..(iii)

From (i), (ii) and (iii), we get

\(\sqrt{3}\) (h – 9) = \(\frac{h-3}{\sqrt{3}}\)

\(\Rightarrow\)  3h -27 = h – 3

\(\Rightarrow\)   3h – h = 27 – 3

\(\Rightarrow\)   2h = 24

\(\Rightarrow\)    h = 12

∴ height of the balloon above the ground is 12 m.

 



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2 Comments

  1. Omparkash Omparkash
    May 27, 2018    

    Exilent

  2. RUBIYA MUSKAAN RUBIYA MUSKAAN
    May 27, 2018    

    Thank u so much rajesh sir I just hope I’ll get the ansers very soon

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