Directions: Solve the following questions applying the various surds and indices formulas.

Q.1.A tower standing on a horizontal plane makes an angle α at a point which is 160m apart from the foot of the tower. On moving 100m towards the base of the tower, the angle of elevation becomes 2α. Find the height of the tower.

a) 100m

b) 80m

c) 120m

d) 90m

Sol : Option B Explanation:
Let AB be the tower of height x m.
Let ∠BCA = α and ∠BDA = 2α such that CD = 100m and AD = 60m

In ΔABD, tan 2α = AB/AD x/60 -------(1)
In Δ ACB, tanα = x/160 -------(2)
Now tan2α = (2tanα)/1-tan^{2}α
(Using (1) & (2))

⇒ 160^{2} – x^{2} = 2 × 160 × 60
⇒ 25600 – x^{2} = 19200
⇒ x^{2} = 6400 ⇒ x = 80
So the height of the tower is 80m

Q.2. A man is standing on a corner of a square field. He observes that the angle of elevation of the top of a pole in the corner just diagonally opposite to this corner is 60^{0}. When he moves 100√2m from the corner along the same straight line (along the diagonal joining the two corners in question), he finds that the angle is 30^{0}. Find the area of the field.

a) 100m^{2}

b) 2500m^{2}

c) 1600m^{2}

d) 1200m^{2}

Sol : Option B Explanation:
Let ABCD is the square field, the person is standing at A and the tower is CE.
Here ∠CAE = 60^{0}
Let the person moves to the point F such that AF = 100√2m and ∠CFE = 30^{0}
Now in ΔACE tan 60^{0} = CE/AC ⇒ CE = √3AC ----(1)
In ΔECF, tan 30^{0} = EC/CF

-----(2)
Equating (1) & (2) we get

⇒ 3AC = 100√2m + AC
⇒ 2AC = 100√2m
⇒ AC = 50√2m
Since AC is the diagonal of the square, so the side of the square is 50m. Hence the area of the square field is (50)^{2} = 2500m^{2}.

Q.3. From the top of a hill 120m high, the angle of depression of top and bottom of a tower are 45^{0} and 60^{0} respectively, find the height of the tower.

a) 40.83m

b) 63.12m

c) 77.14m

d) 50.72m

Sol : Option D Explanation:
Let AB is the hill and CD is the tower.
In ΔABC, tan 60^{0} = AB/AC ⇒√3 = 120/AC ⇒ AC = 120/√3 = 40√3

Now AC = DE = 40√3m
In ΔBED, tan 45^{0} = BE/ED ⇒ 1 ⇒ BE = ED
∴ BE = 40√3m
∴ AE = 120 - 40√3 = 120 – 69.28 = 50.72m
The height of the tower = 50.72m

Q.4.Two boats are sailing in the sea on the two opposite sides of a light house 50m high. The angles of elevation of the top of the light house observed from the ships are 30^{0} and 45^{0} respectively. Find the distance between the two ships.

a) 136.6m

b) 142.8m

c) 150.4m

d) 122.8m

Sol : Option A Explanation:
: Let AB is the light house of height 50m and the ships are at C & D making angles 30^{0} and 45^{0} respectively.
In ΔABC, tan 30^{0} = AB/BC ⇒ 1/√3 = 50/BC ⇒ BC = 50√3m
In ΔABD, tan 45^{0} = AB/BD ⇒ 1 = 50/BD = 50m

∴ The distance between the ships = CD = CB + BD = 50√3 + 50 = 50 ( √3+1)m
= 50 × 2.732 = 136.6m

Q.5.A man 1.5m tall is 10√3m away from the pole. The angle of elevation from his eye to the top of the tower is 60^{0}. Find the height of the tower.

a) 31.5m

b) 36.7m

c) 24.8m

d) 28.6m

Sol : Option A Explanation:
Let AB is the man and CD is the tower.

Here ∠DBE = 60^{0}
In ΔBED, tan 60^{0} = DE/BE
⇒ √3 = DE/10√3
⇒ DE = 10√3 × √3 = 30m
So the height of the tower is DC
= DE + EC = 30 + 1.5 = 31.5m

Q6.When the sun’s altitude changes from 30^{0} to 60^{0}, the length of the shadow of a building decreases by 40m. What is the height of the building?

a) 22√3

b) 15√3

c) 16√2

d) 20√3

Sol : Option D Explanation:
Let AB is the building of height x and the shadow of the tree changed from AC to AD where CD = 40m
In ΔABD, tan 60^{0} = AB/AD

⇒ √3 = x/AD ⇒ AD = x/√3
In ΔACB, tan 30^{0} = AB/AC

⇒ 1/√3 (x/√3 + 40) = x
⇒ x/3 + 40/√3 = x ⇒ 40/√3 = x -x/3
⇒ 40/√3 = 2x/3
⇒ x = (40×3)/2√3 = 20√3m

Q.7.A man standing at a point A, observes that the angles of elevation of two planes one vertically above the other are 60^{0} and 30^{0}. If the higher plane is 1200m from the ground, find the vertical distance between the two planes.

a) 650m

b) 800m

c) 900√3m

d) 1000√3m

Sol : Option B Explanation:
Let the planes are at points C and D and ∠CAB = 30^{0}, ∠DAB = 60^{0}

In ΔABD, tan 60^{0} = BD/AB ⇒ √3 = 1200/AB ⇒ AB = 1200/√3 = 400√3m
In ΔABC, tan 30^{0} = BC/AB ⇒ 1/√3 = BC/400√3 ⇒ BC=400m
∴ The distance between the planes = CD = 1200 – 400 = 800m.

Q8.A tower is surmounted by a vertical flag post of height 20m. At a point on the ground, the angle of elevation of the bottom and the top of the flag post are 30^{0} and 60^{0} respectively. Find the height of the tower.

a) 10m

b) 12m

c) 16m

d) 18m

Sol : Option A Explanation:
Let AB is the tower of height x m and BC is the flag post of height 20m.
In ΔABD, tan 30^{0} = AB/AD ⇒ 1/√3 = x/AD ⇒ AD = √3x ---(1)
In ΔADC, tan 60^{0} = AC/AD ⇒ √3 = (x + 20)/AD ⇒ AD = (x + 20)/√3 ---(2)

From (1) & (2) we have
(x + 20)/√3 = √3x
⇒ x + 20 = 3x
⇒ 2x = 20
⇒ x = 10m
∴ The height of the tower is 10m.

Q9.A balloon leaves from a point A vertically upward at uniform speed. A man standing 200m away from point A observes the angle of elevation of the balloon to be 60^{0} after 3 minutes. Find the speed of the balloon.

a) 2.423m/s

b) 1.924m/s

c) 1.863m/s

d) 2.875m/s

Sol : Option B Explanation:
Let the man is standing at point C, so ∠ACB = 60^{0}.
In ΔACB, tan 60^{0} = AB/AC ⇒ √3 = AB/200

⇒ AB = 200√3 = 200 × 1.732 = 346.4m
∴ The balloon covered 346.4m in 180 sec.
∴ The speed of balloon = 346.4/180 = 1.924 m/s

Q10.The angles of depression and elevation of the top of a building 15m high from the top and bottom of a tower are 60^{0} and 30^{0} respectively. Find the height of the tower

a) 40m

b) 50m

c) 60m

d) 80m

Sol : Option C Explanation:
Let AB is the tower and CD is the building of height 15m
∴CD = AE = 15m
In ΔACD, tan 30^{0} = CD/AC ⇒ 1/√3 = 15/AC

⇒ AC = 15√3m
∴ DE = 15√3m
Again in ΔBED, tan 60^{0} = BE/ED
⇒ √3 = BE/15√3
⇒ BE = 45m
The height of the tower = AB = AE + EB = 15 + 45
= 60m