Height and Distance

itroduction 1. Angle

I f a straight line OA rotates about the point ' O ' called r-e vertex from its initial position to the new position O A ' . Then the angle A O A ' , denoted as Z A O A ' , is formed. The aigle may be positive or negative depending up on their rotation. I f the straight line rotates in anticlockwise direction i positive angle is formed and i f it rotates in clockwise direc--on a negative angle is formed. A n angle is measured in :egree (°). 1 Quadrants

Let X ' O X and Y O Y ' be two lines perpendicular to :h other. The point ' O ' is called the origin, the line X ' O X is lied X axis and Y O Y ' is called the Y axis. These two lines »ide the plane into 4 parts. Each part is called a Q U A D -W T . The part XOY, Y O X ' , X ' O Y ' and Y ' O X are respec-s\y known as 1st, 2nd, 3rd and 4th quadrants. Angle of Elevation

I f an object A ' is above the horizontal line OA we ave to move our eyes in upward direction through an angle

KOA' then the angle A O A ' is called the angle of elevation. I. Angle of Depression

I f an object O is below the horizontal line A ' O ' and *e are standing on the point A ' then we have to move our i>es in downward direction through an angle O ' A ' O . This ingle O ' A ' O is called the angle of depression. 5. Trigonometric Ratio

Let A B C be a right angled triangle. Also let the length : : the sides BC, AC, and A B be a, b and c respectively. Then

A C 1) The ratio

perpendicular b . „ • = — = sm6

BC 2) The ratio

A C 3) The ratio — -

\nd also remember that

1

sin0

hypotenuse

base a = — = cosS

hypotenuse c

perpendicular _ b _

base a

(Hi) c o t e : t a n 0

( iv) t a n 0 = sin 9

cos 6

. COS0 (v) cot9 = 1 ' sin 9 (vi) cos 2 9 + s i n 2 9 = 1

(vii) 1 + tan 2 9 = sec 2 9 (viii) cot 2 9 +1 = cosec 29

6. Values of the trigonometric ratios for some useful angles

4- Ratio/Angle(9)-+ 0° 30° 45° 60° 90" sine 0 1

2 1

& S 2

1

cos 6 1. s 2

1 V2

1 2

0

tan 6 0 1 5

1 s 00 -

sec 9 1 2 73

J5 ' ''2'' CO

cosec6 2 42 2 ^ s

L i

cote CO 1 1 r 0

Rule 1 Problems Based on Pythagoras Theorem

Phythagoras Theorem => h2 = p2 +b2 (see the figure)

(i) cosec0 = (ii) sec 9 =

Illustrative Example Ex: The father watches his son flying a kite from a dis­

tance o f 80 metres. The kite is at a height o f 150 metres directly above the son. How far is the kite from the father?

Soln: Distance o f the kite from the father = FK

COS0

624 P R A C T I C E B O O K ON Q U I C K E R MATHS

(FKf=(FSf+{SKf

[From the above theorem]

.-. FK = V ( l 5 0 ) 2 + (80 ) 2 = 170 metres.

I * 9 mm

Exercise 1. The father watches his son f ly ing a kite from a distance

o f 3 km. The kite is at a height o f 4 k m directly above the son. How far is the kite f rom the father? a) 5 km b) 1 k m c) 7 k m d) None o f these

2. The father watches his son f ly ing a kite from a distance o f 10 metres. The kite is at a height o f 24 metres directly above the son. H o w far is the kite f rom the father? a) 26 m b ) 2 8 m c ) 2 5 m d) Data inadequate

Answers l . a 2.a

Rule 2 Theorem: A man wishes to find the height of a flagpost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagpost to

be 9, . On walking x units towards the tower he finds the

corresponding angle of elevation to be 9 2 * • Then the height

x tan 9, t an0 , (H) of the flagpost is given by t a n 9 2 - t a n 0 , units and

the value of DB (See the figure given below) is given by

.tan 9,

t an0 , - t a n 0 units.

Illustrative Example Ex: A man wishes to f ind the height o f a flagpost which

stands on a horizontal plane; at a point on this plane he finds the angle o f elevation o f the top o f the flagpost to be 45°. On walking 30 metres towards the tower he finds the corresponding angle o f elevation

to be 60°. Find the height o f the flagpost. Soln: Detail Method: A B = height o f flagpost = x m

In AABD

tan 60° = AB

BD

BD s ....(i)

tan 4 5 ° = -AB

BD + DC + 30 = x

V3" = 30

30V3 7 , x = » 7lm

0.732 Quicker Method: App ly ing the above theorem, we have

the required height o f the flagpost

30 x tan 45° x tan 60° tan 6 0 ° - t a n 45°

3 0 x ^ 3 x 1 30V3 « 7 1 m.

V J - l 0.732

Note: 1. The angle o f elevation o f a lamppost changes from

9, to 9 2 when a man walks towards it . I f the height

o f the lamppost is H metres, then the distance trav-

7 / ( t an9 2 - tan 9 ^ elled by man is given by tan 9,. tan 9 2

metres.

2. I f the time for which man walks towards lamppost is given as ' t ' sec then speed o f the man can be calcu­lated by the formula given below.

Speed o f the man = H t a n 9 2 - tan 9,

t tan 9,. tan 9 2

m/sec

Ex: The angle o f elevation o f a lamppost changes from 30° to 60° when a man walks towards it . I f the height

o f the lamppost is l oV3 metres, f ind the distance

travelled by man. Soln: App ly ing the above theorem, we have

the distance travelled by m a n :

r-f 1 A

10V3 V J - ^ =

V J x - L = 20 metres.

Exercise 1. The angle o f elevation o f a lamppost changes from 30"

to 60° when a man walks 20 m towards it . What is the height o f the lamppost?

636 PRACTICE BOOK ON QUICKER MATHS

b; Hint:

P f ' f d m a r k ) R iver Bank

R iver Bank

1 km Required width (PO)

1 x tan 45° x tan 45° l x l x l 1 — gas — lcTT)

• i 45° 4 tan 45° 1 + 1 2 3. d; Hint: Required dis.zuce (x)

"tan 45° +tan 60°

tan 45° x tan 60°

B

*40

40 n \

/ V 5 ° 6 0 ° \

x

' l + V T

4. a; Hint: Required height ( h ) !

B

40 * 63 m

100 x tan 45° x tan 30°

tan 45° + tan 30°

= 5o(V3-l) = 50x0.732 = 36.6 m

Rule 11 Thoerem: From the top and bottom of a building of height

h units, the angles of elevation of the top of a tower are a

and P respectively, then the (i) height of the tower is given

/?tanp by

tan P - tan a units, (ii) distance between the building

and the tower is given by tan P - tan a y

/itana

units and (Hi) RM

units. (Seetheflgure)-^_xma

Note: I f height o f the tower is given as 'H ' units, then the

distance between the building and the tower is giver.

H by _ ~ r7 units.

3 tanp

R •

I I

Illustrative Example Ex: From the top and bottom o f a building o f height 11'.

metres, the angles o f elevation o f the top o f a towe-

are 30° and 45° respectively. Find the height o f the

tower.

Soln: Apply ing the above theorem, we have

120 x tan 45° the height o f the tower =

tan 4 5 ° - t a n 30°

120

1 -fi

120x1.732

.732 * 284 metres

Exercise 1. A tower is 30 m high. An observer from the top of tr*

tower makes an angle o f depression o f 60° at the base I

a building and angle o f depression o f 45° at the top o f

the building, what is the height o f the building? A > :

find the distance between building and tower.

10 a) 12.6m, ^ j m b) 12.6m, 17.3m

c) 12 m, io-y/3 m d) Data inadequate

2. The top o f a 15 metre-high tower makes an angle of el­

evation o f 60° wi th the bottom o f an electric pole anc x angle o f elevation o f 30° wi th the top o f the pole. W :m

is the height o f the electric pole?

[SBI Bank PO Exam, 1 fA

a ) 5 m b ) 8 m c )10m d )12m

3. From the top o f a building 30 m high, the top and bon:

o f a tower are observed to have angles o f depress m

30° and 45° respectively. The height o f the tower •

640 PRACTICE BOOK ON QUICKER MATHS

Then, ZDOB = 3 0 ° ,

ZDOA = 45° &AC = BD = 3000 m.

Let AB = h.

. : — = cot 30° = S =>OD = (3000 x S) DB

m

— = cot 45° = 1 => OC = 3000 m

AC

Distance covered in 15 sec = AB = CD = OD-OC

= (3000V3 - 3000) M = 2196 m

.-. Speed o f the plane

'2196 1 ) Q x 60 x 60 I k m / h r = 527 km/hr

10. d; Hint: Let AB be the c l i f f and C and D be the two positions o f the fishing trawler.

Then, ZACB = 30° and ZADB = 60°

Let AB = h

h

And, — = col30° = fi => AC = fih AB

AD Now, —r- = c o t 6 0 ° =

f CD = AC- AD = fih-

2h JL

Let u m/min be the uniform speed o f the trawler.

Distance covered in 6 min = 611 metres.

2h :. CD = 6u : 6u => h = 3A/3!(

, n A 3V3 w . Now, A D = ~ ^ = = 3 w

Time taken by trawler to reach A

Distance AD 3u 3 m i n .

speed u

11. c; Hint: Let AB be the tower and BC the flagstaff.

Then, BC = —h • Let O be the observer. 5

Then, ZAOC = 45° and Z/4<3B = 9 .

0/4 Now, — = c o t 4 5 ° = l

h + -h 5

• 0 4 = - / j 5

.-. tan 6 = AB h 5

OA ~ 6 ^ " 6

12. d; Hint: Let C be the cloud and C'be its reflection in th lake.

C

AB Now, - ^ - = t a n 3 0 ° = - = r = > x - 2 0 0 = — =

AB VJ VJ

Also, = t a n 6 0 ° = V3

x + 200 = (/15)V3

V3(x-200) =

.-. C5 = 400 m

x + 200 or, x = 400