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

Q.1.A man is standing on the deck of a ship, which is 10m above water level. He observes the angle of elevation of the top of a light house as 60^{0} and the angle of depression of the base of lighthouse as 30^{0}. Find the height of the light house.

a) 30m

b) 40m

c) 45m

d) 38m

Sol : Option B Explanation:
: Let AB is the light house and the man is standing at C so, ∠BCD = 60^{0} and ∠ACD = 30^{0}.
Let BD = h

In ΔADC, tan 30^{0} = 10/CD
⇒ 1/√3 = 10/CD ⇒ CD = 10√3m
In ΔBDC, tan 60^{0} = h/CD
⇒√3 = h/10√3
⇒ h = 30m
So the height of the light house is AB = AD + BD = 10 + 30 = 40m

Q.2. A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 45^{0}. When he moves 20m away from the bank, he finds the angle of elevation to be 30^{0}. Find the height of the tree.

a) 10(√3 + 1)m

b) 15√3m

c) 200(√3 + 1)m

d) 10(√3 - 1)m

Sol : Option A Explanation:
Let AB = x is the tree and AC = y is the river. Let the angle of elevation at point C is 45^{0} and at point D is 30^{0} s.t. CD = 20 m
In ΔACB

tan 45^{0} = x/y ⇒ 1 = x/y ⇒ x = y ……(1)
In ΔADB, tan 30^{0} = AB/AD
⇒ 1/√3 = x/ (20 + y)
⇒ 1/√3 = x/ (20 + x) [ ∵ of (1)]
⇒ 20 + x = √3x ⇒ (√3-1)x = 20
⇒ x = 20/(√3 - 1) = 20/(√3 - 1) x (√3 + 1)/(√3 + 1) = [20(√3 + 1)]/3-1 ⇒ x = [20(√3 + 1)]/2
⇒ x = 10(√3 + 1)m

Q.3. From the top of a building 60m high, the angle of elevation and depression of the top and the foot of another building are α and β respectively. Find the height of the second building.

a) 60(1+ tan α tanβ)

b) 60(1+ cot α tanβ)

c) 60(1+ tan α cotβ)

d) 60(1- tan αcotβ)

Sol : Option C Explanation:
Let AB is the building of height 60m and CD is the second building such that ∠DBE = α and ∠CBE = ∠BCA =β.

In ΔBAC, tanβ = 60/AC
⇒ BE = AC = 60/tanβ = 60cotβ
In ΔBED, tan α = DE/BE ⇒ tanα = DE/60cotβ
⇒ DE = 60 cot β tan α
∴ The height of the building = CD = CE + ED
= 60 + 60 cot β tan α
= 60 (1 + tan α cot β)

Q.4.From the top of a tower 75m high, the angles of depression of the top and bottom of a pole standing on the same plane as the tower are observed to be 30^{0} and 45^{0} respectively. Find the height of the pole.

a) 30.4m

b) 35.9m

c) 28.6m

d) 31.7m

Sol : Option D Explanation:
Let AB is the tower of height 75 m and CD is the pole, such that ∠BDE = 30^{0} and ∠BCA = 45^{0}
In ΔBAC, tan 45^{0} = AB/AC

⇒ 1 = AB/AC ⇒ AB = AC ⇒ AC = 75m
Now DE = AC = 75m
In ΔBED,
tan 30^{0} = BE/DE
⇒ 1/√3 = BE/75 ⇒ BE = 75/√3m
⇒ BE = 25√3m = 43.3 m
Hence the height of the pole
= CD = AE = AB – BE = 75 – 43.3 = 31.7m

Q.5.A 10 m long flagstaff is fixed on the top of a tower on the horizontal plane. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 60^{0} and 45^{0} respectively. Find the height of the tower.

a) 5(√3 + 1)m

b) 5(√3 + 3)m

c) 10(√3 - 1)m

d) 10(√3 + 1)m

Sol : Option A Explanation:
Let AB is the tower of height x m and BC is the flagstaff of height 10 m. Let D be the point from where the angles of elevation are 45^{0} and
60^{0} such that ∠BDA = 45^{0} and ∠CDA = 60^{0}

In ΔDAB, tan 45^{0} = AB/AD
⇒ 1 = x/AD ⇒ AD = x
In ΔDAC, tan 60^{0} = AC/AD
⇒ √3 = (10+x)/x
⇒ √3x = 10+x
⇒ (√3-1)x = 10
⇒ x = 10/(√3 - 1) = 10/(√3 - 1) x (√3 + 1)/(√3 + 1)
⇒ x = 5(√3 + 1)

Q6.The angles of elevation of the top of a tower from two points on the same side of the tower are α and β (α >β). If the distance between the two points is 40m, find the height of the tower.

a) 40cotαcotβ/(tanα + tanβ)

b) 40cotαtanβ/(tanα - tanβ)

c) 40tanαtanβ/(tanα - tanβ)

d) 40tanαtanβ/(tanα + tanβ)

Sol : Option C Explanation:
Let AB is the tower of height x m and C, D are the points where the angles of elevation of the top of the tower are β and α respectively.
Also CD = 40 m

In ΔABD, tan α = AB/AD ⇒ tanα = x/AD ⇒ AD = x/tanα
In Δ ACB, tan β = AB/AC ⇒ tanβ = x/[40 + (x/tanα)]
⇒ x = 40tanβ + x(tanβ/tanα)
⇒ x(1-tanβ/tanα) = 40tanβ
⇒ x[(tanα-tanβ)/tanα] = 40tanβ
⇒ x = 40tanαtanβ/(tanα - tanβ)

Q.7.The angle of elevation of the top of a tower from point A on the ground is 30^{0}. On moving a distance of 40m towards the foot of the tower, the angle of elevation increases to 45^{0}. Find the height of the tower.

a) 48.6m

b) 42.84m

c) 54.64m

d) 58.76m

Sol : Option C Explanation:
Let CD is the tower of height ‘x’ m and A, B are the points where the angles of elevation are 30^{0} and 45^{0} respect.
In Δ BCD, tan 45^{0} = DC/BC

⇒ 1 = x/BC ⇒ BC = x ………(1)
In ΔACD, tan 30^{0} = CD/AC
⇒ 1/√3 = x/(40+x)
⇒ 40+x = √3x
⇒ (√3 - 1)x = 40 ⇒ x = 40/(√3-1) = 40/(√3-1) x (√3+1)/(√3+1) ⇒ x = 20 (√ + 1) = 54.64 m

Q8.An aeroplane, when 4000m high from the ground, pass vertically above another aeroplane at an instance when the angles of elevation of the two aeroplanes from the same point on the ground are 60^{0} and 30^{0} respectively. Find the vertical distance between the two aero planes.

a) 8000/3m

b) 8000/7m

c) 6000/7m

d) 1200m

Sol : Option A Explanation:
Let A and B is the position of the aero planes such that AB = x. Also D is the point on the ground such that ∠BDC = 30^{0} and ∠ADC =60^{0}

In Δ ACD, tan 60^{0} = AC/CD
⇒ √3 = 4000/CD m
⇒ CD = 4000/√3 m
In Δ BCD, tan 30^{0} = BC/CD
⇒ 1/√3 = BC/(4000/√3) ⇒ BC = 4000/√3 x 1/√3 = 4000/3m
∴ The distance between the planes = AB = AC – BC = 4000 - 4000/3 = 8000/3m

Q9.A car is moving at uniform speed towards a tower. It takes 15 minutes for the angle of depression from the top of tower to the car to change from 30^{0} to 60^{0}. What time after this, the car will reach the base of the tower?

a) 6 min

b) 6.5 min

c) 7 min

d) 7.5 min

Sol : Option D Explanation:
Let AB is the tower of height x m. Let C and D be the points on the ground where the angles of depression are 30^{0} and 60^{0} respectively. It took the car 15 minutes to go from C to D.

In Δ ABD, tan 60^{0} = AB/AD ⇒ √3 = x/AD ⇒ AD = x/√3 m
Again in Δ BAC,
tan 30^{0} = x/AC ⇒ 1/√3 = x/AC ⇒ AC = √3x m
Now CD = AC – AD
CD = √3x - x/√3 = 2x/√3 m.
Now the car covered 2x/√3 m in 15 minutes
So it will cover AD = x/√m in 15 × √3/2x × x/√3 = 7.5 minutes.

Q10.A man is watching from the top of a tower, a boat speeding away from the tower. The angle of depression from the top of the tower to the boat is 60^{0} when the boat is 80m from the tower. After 10 seconds, the angle of depression becomes 30^{0}. What is the speed of the boat? (Assume that the boat is running in still water).

a) 20 m/sec

b) 10 m/sec

c) 16 m/sec

d) 18 m/sec

Sol : Option C Explanation:
Let AB is the tower and boat is at points C and D when the angles of depression are 60^{0} and 30^{0} respectively.

In ΔABC,
tan 60^{0}= AB/AC ⇒ √3 = AB/80 ⇒ AB = 80√3m
Again in ΔBAD,
tan 30^{0} = AB/AD ⇒ 1/√3 = 80√3/AD
⇒ AD = 80√3 × √3 = 240m
∴ CD = 240 – 80 = 160m
The boat took 10 seconds to cover 160m
∴ The speed of the boat = 160/10 = 16m/s