Some Applications of Trigonometry Class 10 Notes Mathematics Chapter 9

Heights and Distances

Introduction

We have already mentioned  that trigonometry has a lot of practical applications. One of the most important applications of trigonometry lies in the determination of heights and distances of distant objects, which cannot be measured directly.

The measurement of the heights and distances of objects can be done by observing the angles subtended by these objects at the eyes of the observer. These angles are measured by instruments called Sextants or Theodolites or by other angle-measuring instruments. Knowing these angles, we can then use our knowledge of trigonometric ratios to find these heights and distances.

Angles of Elevation and Angles of Depression

In the discussion of various problems, we shall very often use two terms known as an angle of elevation and angle of depression of an object. These two terms are explained below.

Angle of Elevation :

Let E be the eye of an observer and O be an object above the horizontal line EP through the eye E (Figure).  XY  is the horizontal ground. Then, the line-segment EO joining the eye E and the object O is called the line of sight (or line of vision). The angle of elevation of an object is the angle made by the line of sight with the horizontal through the eye of the observer. In Figure,∠OEP is the angle of elevation of the object O at the observer’s eye E.

Figure

Angle of Depression                                  

Let the observer stand at the top M of a tower or a building and he or she is observing an object O on the ground XY through his or her eye E [Figure]. In this case, since the object lies below the horizontal line through the eye of the observer. Hence, he or she will have to move his or her head downwards and so the observer’s eye will move downwards through an angle starting from the horizontal line EP. This angle is called the angle of depression of the object from the eye. Hence, the angle of depression of an object is the angle made by the line of sight with the horizontal through the eye of the observer. In figure, ∠OEP is the angle of depression of the object O from the observer’s eye E.

Figure

Illustrations

Ex.1.      A tower is100\sqrt 3 metres high. Find the angle of elevation of its top from a point 100 metres away from its foot.

Sol.         Let AB be the tower of height100\sqrt 3 metres and let C be point at a distance of 100 metres from the foot of the tower.

Let q be the angle of elevation of the top of the tower from point C.

In ΔCAB, we have

tan θ =\frac{{AB}}{{AC}}

⇒   tan θ =\frac{{100\sqrt 3 }}{{100}} =\sqrt 3

⇒   θ = 60°

Figure

Hence, the angle of elevation of the top of the tower from a point 100 metres away from its foot is 60°.

Ex.2.      The angles of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it, are complementary. Prove that the height of the tower is\sqrt {ab} metres.

Sol.       Let AB be the tower. Let C and D be two points at distances a and b respectively from the base of the tower. Then, AC = a and AD = b. Let ∠ACB = θ and ∠ADB = 90°– θ

Let h be the height of the tower AB.

In ΔACB, we have tan θ =\frac{{AB}}{{AC}}

⇒   tan θ =\frac{h}{a}   …… (i)

In ΔADB, we have  tan (90° – θ) =\frac{{AB}}{{AD}}

⇒   cot θ =\frac{h}{b}   …… (ii)

Figure

From (i) and (ii), we have

tan θ.cot θ =\frac{{{h^2}}}{{ab}}

⇒  1 =\frac{{{h^2}}}{{ab}}  ⇒ h² = ab  ⇒ h =\sqrt {ab} metres.

Ex.3.      At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the angle of elevation is 3/4. Find the height of the tower.

Sol.         Let AB be the tower and let the angle of elevation of its top at C be α. Let D be a point at a distance of 192 metres from C such that the angle of elevation of the top of the tower at D be β.

Let h be the height of the tower and AD = x.

It is given that

tan α =\frac{5}{{12}}   and  tan β =\frac{3}{4}

In ΔABC, we have tan α =\frac{{AB}}{{AC}}

Figure

⇒   tan α =\frac{h}{{192 + x}}

⇒   \frac{5}{{12}} =\frac{h}{{192 + x}}   … (i)

In Δ ABD, we have

tan β =\frac{{AB}}{{AD}}

⇒  tan β =\frac{h}{x}

⇒   \frac{3}{4} =\frac{h}{x}              … (ii)

We have to find h. This means that we have to eliminate x from equations (i) and (ii).

From equation (ii), we have 3x = 4h  ⇒ x =\frac{{4h}}{3}

Substituting this value of x in equation (i), we get

\frac{5}{{12}} =\frac{h}{{192 + 4h/3}}

⇒   5\left( {192 + \frac{{4h}}{3}} \right)= 12h

⇒   5(576 + 4h) = 36h

⇒   2880 + 20h = 36h

⇒   16h = 2880

⇒   h =\frac{{2880}}{{16}} = 180

Hence, the height of the tower is 180 metres.

Ex.4.      From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 30° and 45°. Find the height of the hill.

Sol.         Let AB be the hill of height h km. Let C and D be two stones due east of the hill at a distance of 1 km from each other such that the angles of depression of C and D be 45° and 30° respectively. Let AC = x km.

Figure

In Δ ABC, we have tan 45°=\frac{{AB}}{{AC}}

⇒   1=\frac{h}{x}  ⇒  h = x           … (i)

In Δ ABD, we have tan 30°=\frac{{AB}}{{AD}}

⇒  \frac{1}{{\sqrt 3 }} =\frac{h}{{x + 1}}

⇒  \sqrt 3 h = x + 1     … (ii)

Substituting the value of x from equation (i) in equation (ii), we get

\sqrt 3 h = h + 1  ⇒ h (\sqrt 3 – 1) = 1

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

=\frac{{\sqrt 3  + 1}}{2} =\frac{{2.73}}{2} = 1.365

Hence, the height of the hill is 1.365 km.

Ex.5.      Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If α, β be the elevations of the top of the tower from these stations, prove that its inclination θ to the horizontal is given by

cot θ =\frac{{b\,\cot \alpha  - a\,\cot \beta }}{{b - a}}

Sol.    Let AB be the leaning tower and let C and D be two given stations at distances a and b respectively from the foot A of the tower.

Figure

Let AE = x and BE = h

In Δ ABE, we have tan θ =\frac{{BE}}{{AE}}

⇒   tan θ =\frac{h}{x}

⇒   x = h cot θ                                                                    … (i)

In ΔCBE, we have tan a=\frac{{BE}}{{CE}}

⇒   tan α =\frac{h}{{a + x}}

⇒  a + x = h cot α

⇒   x = h cot α – a                           … (ii)

In ΔDBE, we have tan β =\frac{{BE}}{{DE}}

⇒   tan β =\frac{h}{{b + x}}

⇒   b + x = h cot β

⇒   x = h cot β – b                                      … (iii)

From equations (i) and (ii), we have

h cot θ = h cot α – a

⇒    h (cot α – cot θ ) = a

⇒   h =\frac{a}{{\cot \alpha  - \cot \theta }}                     … (iv)

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

h cot θ = h cot β – b

⇒  h (cot β – cot θ) = b

⇒   h = \frac{b}{{\cot \beta  - \cot \theta }}                   … (v)

Equating the values of h from equations (iv) and (v), we get

\frac{a}{{\cot \alpha  - \cot \theta }}=\frac{b}{{\cot \beta  - \cot \theta }}

⇒   a (cot β – cot θ) = b (cot α – cot θ)

⇒   (b – a) cot θ = b cot α – a cot β

⇒   cot θ =\frac{{b\,\cot \alpha  - a\,\cot \beta }}{{b - a}}

Ex.6.      There are two pillars of equal height and on either side of a road, which is 100 m wide. The angles of elevation of the top of the pillars are 60° and 30° at a point on the road between the pillars. Find the position of the point between the pillars and the height of each pillar.

Sol.         Let AB and CD be two pillars, each of height h metres. Let P be a point on the road such that AP = x metres. Then, CP = (100 – x) metres. It is given that ∠APB = 60° and ∠CPD = 30°.

Figure

In ΔAPB, we have tan 60°=\frac{{AB}}{{AP}}

⇒   \sqrt 3 =\frac{h}{x}

⇒   h =\sqrt 3 x                                                   … (i)

In ΔCPD, we have tan 30° =\frac{{CD}}{{CP}}

⇒   \frac{1}{{\sqrt 3 }} =\frac{h}{{100 - x}}

⇒    h\sqrt 3 = 100 – x                         … (ii)

Eliminating h between equations (i) and (ii) we get

3x = 100 – x  ⇒ 4x = 100  ⇒ x = 25

Substituting x = 25 in equation (i), we get

h = 25\sqrt 3 = 25 × 1.732 = 43.3

Thus, the required point is at a distance of 25 metres from the first pillar and 75 metres from the second pillar. The height of the pillar is 43.3 metres.

Ex.7.      The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m verticallyabove X, the angle of elevation is 45°. Find the height of the tower PQ and the distance XQ.

Sol.     In ΔYRQ, we have

tan 45° =\frac{{QR}}{{YR}}

⇒   1 =\frac{x}{{YR}}

⇒   YR = x

⇒   XP = x       [ YR = XP]

In DPQX, we have tan 60° =\frac{{PQ}}{{PX}}               

\sqrt 3 =\frac{{x + 40}}{x}  [ PX = x]

Figure

⇒   \sqrt 3 x = x + 40

⇒   x (\sqrt 3 – 1) = 40

⇒   x =\frac{{40}}{{\sqrt 3  - 1}}

⇒   x =\frac{{40}}{{\sqrt 3  - 1}}  ×\frac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}} = 20(\sqrt 3 + 1) = 54.64

So, height of the tower PQ = x + 40 = 54.64 + 40 = 94.64 metres.

In ΔPQX, we have sin 60°  =\frac{{PQ}}{{XQ}}

⇒   \frac{{\sqrt 3 }}{2} =\frac{{94.64}}{{XQ}}

⇒   XQ =\frac{{94.64 \times 2}}{{\sqrt 3 }}

⇒    XQ = \frac{{94.64 \times 2 \times \sqrt 3 }}{3}= 109.3 metres.

Ex. 8.     The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of cloud in the lake is 60°. Find the height of the cloud.

Sol.         Let AB be the surface of the lake and P be the point of observation such that AP = 60 metres. Let C be the position of the cloud and C’ be its reflection in the lake. Then, CB = C’B. Let PM be perpendicular from P on CB. Then ∠CPM = 30° and ∠C’PM = 60°. Let CM = h. Then CB = h + 60. Consequently, C’B = h + 60.

In ΔCPM, we have

tan 30° =\frac{{CM}}{{PM}}

⇒   \frac{1}{{\sqrt 3 }}=\frac{h}{{PM}}

⇒   PM =\sqrt 3 h                                 … (i)

In ΔPMC’, we have tan 60° =\frac{{C'M}}{{PM}}    

 

Figure

⇒  tan 60° =\frac{{C'B + BM}}{{PM}}

⇒   \sqrt 3 =\frac{{h + 60 + 60}}{{PM}}

⇒  PM =\frac{{h + 120}}{{\sqrt 3 }}                   … (ii)

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

\sqrt 3 h =\frac{{h + 120}}{{\sqrt 3 }}

⇒   3h = h + 120

⇒   2h = 120

⇒  h = 60

Now CB = CM + MB = h + 60 = 60 + 60 = 120

Hence, the height of the cloud from the surface of the lake is 120 metres.

Ex.9.      An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at that instant.

Sol.      Let P and Q be the positions of two aeroplanes where Q is vertically below P and OP = 4000 m. Let the angles of elevation of P and Q at a point A on the ground be 60° and 45° respectively.

In triangles AOP and AOQ, we have      

Figure

tan 60° =\frac{{OP}}{{OA}}  and  tan 45° =\frac{{OQ}}{{OA}}

⇒   \sqrt 3 = \frac{{4000}}{{OA}} and  1 =\frac{{OQ}}{{OA}}

⇒   OA =\frac{{4000}}{{\sqrt 3 }}  and  OQ = OA

⇒   OQ =\frac{{4000}}{{\sqrt 3 }} m

∴  Vertical distance between the aeroplanes = PQ

= OP – OQ

=\left( {4000 - \frac{{4000}}{{\sqrt 3 }}} \right)m = 4000\frac{{(\sqrt 3  - 1)}}{{\sqrt 3 }}m = 1690.53 m

Ex.10.    A tall tree is broken by the wind in such a way that its top touches the ground making an angle of  30° with the ground. The distance from the foot of  the tree to the point where the top touches the ground is 20 metres. Find the height of the tree.   [\sqrt 3 \, = \,1.73]

Sol.       Let AB (= x metres) be the tree.

Suppose the tree breaks at point P and then part AP assumes the position CP, meeting the ground at point C.

Let    PB = y metres

Then  AP = AB – PB = (x – y) metres,

and    PC = AP = (x – y) metres

∠ PCB = 30° and BC = 20 metres

Figure                                                

∠ PBC = 90°

In right ΔPBC, we have     tan 30° = \frac{{{\rm{PB}}}}{{{\rm{BC}}}}

⇒   \frac{1}{{\sqrt 3 }}=\frac{y}{{{\rm{20}}}}

⇒   y =\frac{{20}}{{\sqrt 3 }}              . . .(1)

and sin 30° =\frac{{{\rm{PB}}}}{{{\rm{PC}}}}

⇒   \frac{{\rm{1}}}{{\rm{2}}}=\frac{y}{{x\, - \,y}}

x – y = 2y                                                                                      

x = 3y                  . . .(2)

From (1) and (2), we have x = 3×\frac{{20}}{{\sqrt 3 }} = 20\sqrt 3   = 20(1.73) = 34.6

Hence, the height of the tree is 34.6 m

Ex. 11.   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.

Sol.      Let AB (= h metres) be the tower.

Let C be the initial position of the car on the highway.

Let D be the position of the car after 6 seconds.

Let AH be the horizontal through A.

HA is parallel to CDB.

The angle of depression of point C at A is 30° and the angle of depression of point D at A is 60°.

i.e.  ∠ACB = ∠HAC = 30°

and    ∠ADB = ∠HAD = 60°

∠ABC = ∠ABD = 90°

Let CD = x metres and let DB = y metres (distance to be covered to reach the tower)

In right triangle ABD, we have

tan 60° = \frac{{{\bf{AB}}}}{{{\bf{DB}}}}

Figure

⇒      \sqrt 3 = \frac{h}{y}

⇒      \sqrt 3 y= h          . . .(1)

In right triangle ABC, we have

tan 30° =\frac{{{\bf{AB}}}}{{{\bf{CB}}}}

⇒     \frac{1}{{\sqrt 3 }} = \frac{h}{{{\bf{CD + DB}}}}

⇒     \frac{1}{{\sqrt 3 }} = \frac{{\sqrt 3 y}}{{x + y}}                   [Using (1)]

⇒       x + y = 3y

⇒      x = 2y                       . . .(2)

The car covers x metres in 6 seconds             [Given]

i.e.  the car covers 2y metres in 6 seconds          [Using (2)]

∴    the car covers y metres in 3 seconds.

Hence, the car will reach the foot of the tower in 3 seconds.

Ex. 12.   A man in a boat rowing away from lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top oflighthouse from 60° to 45°. Find the speed of  the boat .[\sqrt 3 \,\, = \,\,1.732]

Sol.         Let AB (=10 m) be the lighthouse. Let C and D be the points of observation. The angle of elevation of the top A of the lighthouse AB at C is 60° and the angle of elevation of the top A of the lighthouse AB at D is 45°.

i.e. ∠ ACB = 60°, ∠ ADB = 45°

∠ ABD = ∠ ABC = 90°

Let BC  = x  metres  and CD = y metres

In right triangles ABC, we have

tan 60° = \frac{{{\rm{AB}}}}{{{\rm{BC}}}}

⇒   \sqrt 3 = \frac{{100}}{x}

⇒   x = \frac{{100}}{{\sqrt 3 }}                                  ….(1)

In right triangle ABC, we have    

Figure

tan 45° = \frac{{{\rm{AB}}}}{{{\rm{BD}}}}

1 =\frac{{100}}{{x + y}}

⇒  x+y = 100                            ….(2)

On substituting the value of x from equation (1) in equation (2) we, get

\frac{{100}}{{\sqrt 3 }}\, + \,y= 100

⇒    y =100\, - \,\frac{{100}}{{\sqrt 3 }}\, = \,100\,\left( {1\, - \,\frac{1}{{\sqrt 3 }}} \right)\,\, = \,\,100\,\left( {\frac{{\sqrt 3 \, - \,1}}{{\sqrt 3 }}} \right)….(3)

In 2 minutes, the distance covered by the boat is y m,

∴   In 120 seconds the distance covered by the boat =\frac{{100\,(\sqrt 3 \,\, - \,1)}}{{\sqrt 3 }} m  [Using (3)]

⇒  In 1 second the distance covered by the boat  =\frac{{100(\sqrt {3\,} \, - \,1)}}{{120\sqrt 3 }}\,\frac{{\sqrt 3 }}{{\sqrt 3 }}\,\,{\rm{m}}

=\frac{{100(\sqrt {3\,} \, - \,1)\sqrt 3 }}{{120 \times \,3}}\,{\rm{m}} =\frac{5}{{18}}\,(3\, - \,\sqrt 3 ){\rm{m}}

∴Speed of the boat = \frac{5}{{18}}\,(3\, - \,\sqrt 3 )\,{\rm{m/s}}

=\frac{5}{{18}}\,(3\, - \,\sqrt 3 ) \times \,\frac{{18}}{5}\,{\rm{km/h}}

=(3\, - \,\sqrt 3 \,)\,{\rm{km/h}}

= (3 – 1.732) km/h

= 1.268 km/h

Hence the speed of the boat is 1.268 km/h 

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