# CLASS-10TRIGONOMETRY - HEIGHT AND DISTANCE - PROBLEM & SOLUTION

There are some important problem & solution has given below to clear concept about Trigonometry - Height & Solution

Example.1) A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60⁰ and the angle of depression of the foot of the tower is 30⁰. Find the height of the tower if the height of the pole is 20 m

Ans.)  Let, AB be the pole and CD be the tower. Draw AE ⊥ CD. Then,

AB = 20 m, ∠CAE = 60⁰ and ∠ADB = 30⁰

Let, BD = x m and CD = h meters

Then, CE = (CD – ED) = (CD – AB) = (h – 20) m

From right △ABD, we have

AB

------- = tan 30⁰

BD

20           1

=>    ------- = -------

x           √3

=>      x  =  20 √3 …………………..……….(i)

Clearly, AE = BD = x meters

From right AEC, we have

CE

-------- =  tan 60⁰

AE

(h – 20)

=>    ----------  =  √3

x

(h – 20)

=>    x  =  ------------  …………………………(ii)

√3

Equating the value of x from (i) and (ii), we get –

(h – 20)

20 √3 =  -----------

√3

=>   (h – 20)  =  (20√3 X √3)

=>      h = (20 + 60) =  80

Hence the height of the tower is 80 m   (Ans.)

Example.2) In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30⁰ and 60⁰ respectively. Find –

(i) The horizontal distance between AB & CD

(ii) The height of the lamp.

Ans.)  Let, AX be the horizontal line

Draw DE ǁ CB. Then,

∠ADE = 30⁰  and  ∠ACB = 60⁰

Let, CB = DE = x meters

And CD = BE = y meters  Then, AE = (60 – y) m

From right △ABC, we have –

AB

-------- =  tan 60⁰

BC

60

=>    -------- =  √3

x

=>        x  =  20 √3   …………………(i)

From right △AED, we have –

AE

--------  =  tan 30⁰

ED

(60 – y)         1

=>    ---------- = ------

x            √3

=>       x  =  (60 – y) √3  …………………………..(ii)

Equating the value of x from (i) and (ii), we get –

20√3 = (60 – y) √3

=>        20√3 = 60√3 - √3y

=>        √3y  = 60√3 - 20√3

=>         √3y = 40√3

=>          y  =  40

Thus the horizontal distance between AB & CD is 20√3 = 34.64 m   …………..(i)     (Ans.)

And, the height of the lamp post is 40 m  ……………………(ii)      (Ans.)

Example.3)  As observed from the top of a 80 m tall light house, the angles of depression of the two ships on the same side of the light house in horizontal line with its base are 30⁰ and 40⁰ respectively. Find the distance between the two ships. Give your answer correct to the nearest meters.

Ans.) Let AB be the lighthouse and C and D be the two ships Let, AX be the horizontal line. Then,

AB = 80 m, ∠ACB = ∠CAX = 30⁰  and ∠ADB = ∠DAX = 40⁰

Let, CD = x meters and DB = y m

From right △ABD, we have

AB

-------- = tan 40⁰

DB

80

=>      ------- =  0.84

y

80

=>        y  =  --------  =  95.24

0.84

From right △ABC, we have –

AB

-------  =  tan 30⁰

CB

80           1

=>   -------- = ------

x + y         √3

=>      (x + y) =  80 √3  = (80 X 1.732)      [where, √3 = 1.732]

=  138.56

Now,   CD  =  x  = (x + y) – y = (138.56 – 95.24)

=   43.32

Hence, the required distance between the two ships is 43 m.   (Ans.)

Example.4) The shadow of a tower, when the angle of elevation of the sun is 45⁰, is found to be 10 m longer than when it was 60⁰. Find the height of the tower.

Ans.)  Let AB be the tower and let AC and AD be the shadows of the tower at the two instants. Then, ∠ACB = 60⁰, ∠ADB = 45⁰, and CD = 10 m.

Let, CA = x meters and AB = h meters

From right △DAB, we have

------- =  cot 45⁰

AB

10 + x

=>       --------- = 1,

h

=>          h  =  (10 + x)

=>          x  =  (h – 10)    ……………………………(i)

Again from right △CAB, we have –

AC

-----------  =  cot 60⁰

AB

x            1

=>    ------- =  -------

h           √3

h

=>     x  =  -------   …………………….(ii)

√3

Equating the value of x from (i) and (ii), we get –

h

(h – 10)  =  -------

√3

=>    (h – 10)√3  =  h

=>    √3 h - 10√3 = h

=>    √3h – h = 10√3

=>    (√3 – 1)h = 10√3

10√3

=>   h  = -----------

(√3 – 1)

10√3             (√3 + 1)

=>   h  = ---------  X  ------------

(√3 – 1)           (√3 + 1)

10√3 (√3 + 1)           10√3 (√3 + 1)

=>   h  =  -----------------  = ----------------

(3 – 1)                        2

=>    h  =   5√3 (√3 + 1)

=>    h  =   {5 X (√3)²} + 5√3

=>    h  =  15 + (5 X 1.732) = (15 + 8.66)

=>    h  =  23.66

Hence the height of the tower is 23.66 m.   (Ans.)