Height & Distance

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ICSE – Mathematics Important Q.A.

Height and Distance.

Q.1. From the top of a hill, the angle of depression of two consecutive kilometer stones, due east are found to be 30º and 45º respectively. Find the distance of the two stones from the foot of the hill. [Use √3 = 1.732]

Solution : –

Fig.

Let AB be the hill whose foot is B and D and C are two kilometer stones.

Therefore  DC = 1 km = 1000 m, Let AB = h and BC = x.

In right angled Δ ABC, tan 45º = AB/BC  =>  1 = h/x  =>  x = h   ————- (1)

In right angled Δ ABC, tan 30º = AB/BD =>  1/√3 = h/(x + 1000)

=> x + 1000 = h√3    —————-(2)

using (1) and (2) we get    x + 1000 = x√3  =>  x(√3 – 1) = 1000

Or,                                x = [1000/(√3 – 1)] × (√3 + 1)/(√3 + 1)

Or,                                x = 1000 (√3 + 1)/2 = 500(√3 + 1) = 500 × 2.732

= 1366 m = 1.366 km.

Therefore first km stone is 1.366 km and second km stone is 2.366 km from the foot of the hill. [Ans.]

Q.2. A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60º. When he moves 50 m away from the bank, he finds the angle of elevation to be 30º. Calculate :

(ii) the width of the river and

(ii) the height of the tree.

Solution : –

Let AD be the tree of height h, In ΔADC, tan60º = h/CD

Or,             √3 = h/CD

Or,             CD = h/√3

In ΔADB,     tan30º = h/BD

Or,  1/√3 = h/BD

Or,    BD = h√3

BD – CD = 50

h√3 – h/√3 = 50

(3h – h)/√3 = 50

2h = 50√3

(i)      height of the tree,     h = 50√3/2 = 25√3 25×1.732 =  43.3 m. [Ans.]

(ii)  Width of the river = CD = h/√3 = 25√3/√3 = 25 m. [Ans.]

Q.3. In the figure (not drawn to scale) TF is a tower. The elevation of T, from A is xº where tan xº = 2/5 and AF = 200 m. The elevation of T from B, where AB = 80 m is yº. Calculate :

(i) the height of the tower TF

(ii) the angle ’y’ correct to nearest degree.

Solution : –

(i) In Δ ATF, tan xº = TF/AF

Or,                    2/5 = TF/200

Or,               5TF = 400

Or,                    TF = 400/5 = 80 m. [Ans.]

(ii)   BF = AF – AB = 200 – 80 = 120 m

In Δ BTF, tan yº = TF/BF = 80/120 = 2/3

Or,           tan yº = 0.6667

From the table we get y = 34º. [Ans.]

Q.4. The angle of elevation of the top of a tower from two points P and Q at a distance of ‘a’ and ‘b’ respectively, from the base and in the same straight line with it are complementary. Prove that height of the tower is √(ab).

Solution : –

Let L SPR = θ then L SQR = 90º – θ and RS = h,

In Δ SPR, tan θ = h/a  Or, h = a tan θ  ———————————  (i)

In ΔSQR, tan (90º – θ) = h/b Or, h = b tan (90º – θ) = h cot θ —–  (ii)

Multiplying (i) and (ii) we get,  h2 = a tan θ b cot θ = a b

Or,   h = √(a b).      [Proved.]

Q.5. From a window A, 10 m above ground the angle of elevation of the top C to a tower is xº , where tan xº = 5/2 and the angle of depression of the foot D of the tower is yº , where tan yº = 1/4 . Calculate the height CD of the tower in metres.

Solution : –

L BDA = L DAE = yº ,             [alternate angle]   and   AB = DE

In Δ ABD, tan yº = AB/BD

Or,             1/4 = 10/BD

Or,              BD = 40 m = AE

In Δ AEC, tan xº = CE/AE

Or,              5/2 = CE/40

Or,               CE = (540)/2 = 200/2 = 100 m,

CD = CE + DE = 100 m + 10 m = 110 m. [Ans.]

Q.6. The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45º to 30º. Find the height of the tower, correct to two decimal places.

Solution : –

Let the height of the tower AB be h and length of the shadow initially BC = y.

In Δ ABC, tan 45º = AB/BC

Or,  1 = h/y

Or,   y = h    —————— (i)

In Δ ABD,  tan30º = AB/BD

Or, 1/√3 = h/(y + 10)

Or, y + 10 = h√3   —————  (ii)

Putting y = h in equation (ii)    h + 10 = h√3

Or,  h (√3 – 1)  = 10

Or,   h = 10/(√3 – 1)  = {10 (√3 + 1)}/{(√3 – 1)(√3 + 1)}

= 10 (√3 + 1)/2

= 5(1.732 + 1)

= 5×2.732 = 13.66 m. [Ans.]

Q.7. From the top of a cliff 92 m high, the angle of depression of a buoy is 20º. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff.

Solution : –

Let C be the buoy and A the cliff, L EBC = 20º = L ACB [alternate angle]

In Δ CAB,       cot 20º = AC/AB

Or,                         2.7475 = AC/92

Or,                                AC = 2.7475 × 92 = 252.77 = 253 m. [Ans.]

Q.8. Two poles standing on the same side of a tower in a straight line with it, measure the angles of elevation of the top of the tower at 25º and 50º respectively. If the height of the tower is 70 m, find the distance between the two people.

Solution : –

Let AB be the tower and C and D the poles. AB = 70 m.

In Δ ABC,  cot 50º = BC/AB

Or,        cot (90º – 40º) = BC/70

Or,                    tan 40º = BC/70

Or,                    0.8391 = BC/70

Or,                          BC = 0.8391 × 70 = 58.74 m.

In Δ ABD,   cot 25º  = BD/AB

Or,      cot (90º – 65º) = BD/70

Or,                  tan 65º = BD/70

Or,              2.114451 = BD/70

Or,                       BD  = 2.114451 × 70 = 148.01 m.

Hence, distance between people = CD = BD – BC

= 148.01 – 58.74 = 89.27 m. [Ans.]

Q.9. Vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?

Solution : –

Fig.

We have, cos θ = 0.53, let distance of the man from the foot of the tower be x.

AB = 20m, BC = x, then AC = √[x2 + (20)2] = √[x2 + 400]

cos θ = BC/AC = x/√[x2 + 400]

Or,                     0.53 = x/√[x2 + 400]

Or,                     (0.53)2 = x2/[x2 + 400] [Squaring both sides]

Or,                    0.2809 = x2/[x2 + 400]

Or,                        x2 = 0.2809x2 + 112.36

Or,               x2 – 0.2809x2 = 112.36

Or,                    0.7191x2 = 112.36

Or,                              x2 = 112.36/0.7191 = 12.5 m [Ans.]

Q.10. With reference to the figure given a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of a vertical pole BC. The height of the pole is 10 m. The man’s eye is 2 m above the ground. He observes the angle of elevation at C, the top of the pole as xº, where tan xº = 2/5. Calculate : (i) The distance AB in metre. (ii) The angle of elevation of the top of the pole when he is standing 15 m from the pole. Give your answer to the nearest degree.

Fig.

Solution : – Do yourself. [AB = 20 m, Angle = 28º.]

Q.11. The figure drawn alongside is not to the scale. AB is a tower, and two objects C and D are located on the ground on the same side of AB. When observed from the top A of the tower, their angles of depression are 45º and 60º. Find the distance between the two objects if the height of the tower is 300m. Give your answer to the nearest metre.

Fig.

Solution : –

Form figure, LXAD = LADB = 60º and LXAC = LACB = 45º.

In right-angled triangle ADB, tan60º = AB/DB

Or,                                               √3 = 300/DB

Or,                                               DB = 300/√3 = 100√3 = 100×1.732 = 173.2 m.

In right-angled triangle ACB,  tan45º = AB/CB

Or,                                                  1 = 300/CB

Or,                                                CB = 300 m.

Hence, the distance between two objects = CD = CB – DB

= 300 – 173.2

= 126.8 = 127 m (approx). [Ans.]


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