MCQ: Height & Distance - 1 - SSC CGL MCQ

Let AB be the man and BC be his shadow

⇒ AB = 180

⇒ tan60° = AB/BC

√3 = 180/BC

∴ BC = 103.92 cm

Let AB be the tree bowed at the point C so that part CB takes the position CD.
Then, CD = CB.
Let AC = x meters. At that point, CD = CB = (10 - X) m and ∠ADC = 60°.
AC/AC = sin60° => x/(10 - x) = √3/2
=> 2x = 10 √3 - √3x
=> (2 + √3) x = 10 √3
=>x = 10 √3/ (2 + √3) * (2 - √3)/(2 - √3) = 20 √3 - 30)m
= (20 * 1.73 - 30) m = 4.6m
=> Required height = 4.6m.

Let AB be the tower and CD be the observer

tan30° = AE/CE

⇒ 1/√3= (AB – CD) / x = (50 – 1.5) / x = 48.5/x

x = 84 m

Let AB be the kite and AC be the level ground
So that BC - AC.
At that point, ∠BAC = 60°and BC = 75m. Let AB = x meters.
Presently AB/BC = coses60° = 2/ √3
=> x/75 = 2/√3 => x = 150/√3 = 150 * √3/3 = 50 √3m.
∴ Length of the string = 50 √3m.

Let AB be the light house and the two boats be at C and D

∴ AB = 125 m

tan30° = BC/AB = x/125 = 1/3

=> x = 72.17 m

tan45° = BD/AB = y/125 = 1

=> y = 125 m

∴ The distance between the two boats is = x + y

= 72.17 + 125

= 197.17 m

Let AB be the post and AC be its shadow.
Let AB = X m. Then, AC = √3Xm.Let∠ACB = θ.
AB/AC = tanθ => tanθ = X/√3X = 1/√3 = tan30°.
∴ θ = 30°.
Hence, the point of rise is 30°.

Let AB be the building and the man is standing at A.

When the car is 200m away from the building, the angle of depression is 60°

tan 60° = BD/AB = √3

⇒ 200/AB = √3

⇒ AB = 115.47 m

⇒ tan30° =BC/AB = x/115.47 = 1/√3

⇒ x = 66.67

Now, the car travels distance CD in 8 seconds

 CD = BD – BC = 200 – 66.67 = 133.33 m

 Speed = 133.33/8 = 16.67 m/s

Let AB be the tower and let C and D be the two's positions men.
At that point ∠ACB = 30°,∠ADB = 45°and AB = 50 m
AC/AB = Cot30° = √3 => AC/50 = √3
=> AC = 50√3m
AD/AB = cot 45° = 1 => AD/50 = 1
=> AD = 50M.
Separation between the two men = CD = (AC + AD)
= (50√3 + 50) m = 50(√3 + 1)
=50(1.73 + 1)m = (50 * 2.73)m = 136.5m.

Let AB be the observation tower and h be its height.

Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.

The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.

∴ tan 30º = AB/CB = h / CB = 1/ √3

⇒ CB = √3 x h

∴ tan 45º = AB / DB = h / DB = 1

⇒ DB = h

⇒ CD = CB – DB = (√3h – h) = h(√3– 1)

Now, as h(√3 – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds

Let AB be the building and AC be its shadow.
Then, AC = 20m and ∠ACB = 60°. Let AB = x m.
Presently AB/AC = tan 60° = √3 => x/10 = √3
=> x = 10√3m = (10 * 1.732) m = 17.32m.
∴ Height of the building is 17.32m.

Let AB be the taller building of height 60 m and CD be the smaller one of height h m.

⇒ DB / AB = tan 30º = 1/√3

⇒ DB = AB x tan 30º

= 34.64 m.

tan 30º = AE/CE = AE/DB = 1/ = AE/34.64

∴ AE = 60 – h = 20

∴ h = 40 m

Let the total length of the tree be X + Y meters
From the figure tan 45 = X/20 => X = 20
cos 45 = 20/Y => Y = 20/cos 45 = 20√2
X + Y = 20 + 20radic; 2 = 20 + 2 x 10 x 1.414 = 48.28 meters

Let AB be the observer and CD be the tower.

Draw BE ⊥ CD.

Then, CE = AB = 1.6 m,

BE = AC = 20√3 m.

∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.

Let C and D be the position of the aeroplanes.
Given that CB = 900 m,∠CAB = 60°,∠DAB = 45°
From the right △ ABC,
Tan45 = CB/AB => CB = AB
From the right △ ADB,
Tan30 = DB/AB => DB = ABtan30 = CBx(1/√3) = 750/√3
CB = CD + DB
=> Required height CD = CB - DB = 750 - 750/√3 = 250(3 - √3)

Let AB be the wall and BC be the ladder.

Then, ACB = 60º and AC = 4.6 m.

⇒ BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.