Heights and Distances MCQ Quiz - Objective Question with Answer for Heights and Distances - Download Free PDF

Solution:

let the distance between the top of the flagstaff and the observer is x m.

In \(\triangle\)OFA,  \(\tan 2α =\frac{30}{x}\) 
\( \because \tan 2 α=\frac{2 \tan α}{1-\tan ^2 α} \)
\( \Rightarrow \frac{2 \tan α}{1-\tan ^2 α}=\frac{30}{x} \ldots (i)\)

In \(\triangle\) OFE, \( \tan α=\frac{5}{x} \ldots(i i)\) 
From (i) and (ii), we get
\( \Rightarrow \frac{2 \times(5 / x)}{1-(5 / x)^2}=\frac{30}{x} \)
\( \Rightarrow \frac{10 x^2}{x\left(x^2-25\right)}=\frac{30}{x} \)
\( \Rightarrow \frac{x^2}{x^2-25}=3 \)
\( \Rightarrow x^2=3 x^2-75 \)
\( \Rightarrow 2 x^2=75 \)
\( \Rightarrow x=\sqrt{\frac{25 \times 3}{2}} \)
\( \Rightarrow x=5 \sqrt{\frac{3}{2}} \)

Refer to the figure. Let the height of the tree be h. Let the part that is still standing on the ground be x, and that part which has fallen be y. Hence, we have

sin θ = x/y

It is given that θ = 30

⇒ sin θ = ½

⇒ x/y = ½

⇒ y = 2x

Also, it is given that the distance of the tip of the fallen tree to that of the base of the tree is 8 m

⇒ cos θ = 8/y

⇒ √3/2 = 8/y

⇒ y = 16√3/3

Hence, x = 8√3/3

h = x + y = 8√3/3 + 16√3/3 = 24√3/3 = 8√3 m

∴ The height of the tree = 8√3 m

Calculation:

After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. 

⇒ v = 432 × 1000/(60 × 60) m/sec

= 120 m/sec

∴ Distance PQ = v × 20 = 2400 m

Now, In ΔPAC

tan 600 = h/AC

⇒ AC = h/(√3)

In ΔAQD

tan 30° = h/AD

⇒ AD = √3h

∴ AD = AC + CD

⇒ √3 h = h/√3 + 2400

⇒ 2h/√3 = 2400

⇒ h = 1200√3 m

∴ The height is 1200√3 m.

The correct answer is Option 1.

Calculation:

Let PQ be the house subtending a right angle at the window B of opposite house AB 

In ΔPAB, we have

\(\tan 60^{\circ}=\frac{A B}{6}=A B=6 √{3} m\) = CP

BC = AP = 6

In ΔCBQ, we have

\(\tan 30^{\circ}=\frac{h-C P}{B C}\)

⇒ \(h=6\left(√{3}+\frac{1}{√{3}}\right)\)

⇒ h = 8√3 m

Hence option 1 is correct

Given:

The angle of depression = 60º

Distance from the base of the tower = 70 meters

Formula Used:

In a right triangle, tan(θ) = Opposite Side / Adjacent Side

Calculation:

Let the height of the tower be h meters.

tan(60º) = h / 70

⇒ √3 = h / 70

⇒ h = 70 × √3

⇒ h = 70√3 meters

The height of the tower is 70√3 meters.

Given:

Height of building = 1000 m

Angle of elevation = 30°

Formula used:

Tan θ = perpendicular/Base = P/B

Tan 30° = 1/√3

Value of √3 = 1.732 

Calculation:

Tan 30° = AB/BC

⇒ 1/√3 = 1000/BC

⇒ BC = 1000√3

⇒ BC = 1000 × 1.732 = 1732 m

∴ The correct answer is 1732 m.

Calcualation:

Here, AD and BE are the same ladders.

So, AD = BE

In Δ ACD,

AD² = AC² + CD² = (y + 0.8)² + 2.5²

In Δ BCE,

BE² = BC²+ CE² = y² + (2.5 + 1.4)² = y² + 3.9²

Since, AD = BE

So, AD² = BE²

(y + 0.8)2 + 2.52 =  y2 + 3.92

⇒ y² + (0.8)² + 2 × y × 0.8  + 6.25 = y² + 15.21

⇒ 0.64 + 1.6y + 6.25 = 15.21

⇒ 1.6y = 15.21 - 0.64 - 6.25 = 8.32

⇒ y = 8.32 / 1.6 = 5.2

So,

AD² =  (y + 0.8)2 + 2.52 = (5.2 + 0.8)² + 2.5² = 36 + 6.25 = 42.25

So, Length of the ladder = AD = √42.25 = 6.5 m

∴ The correct answer is option (2).

Let the height of the platform be x m.

⇒ tan45° = Perpendicular/Base = (47 – x) / 40

⇒ 1 = (47 – x) / 40

⇒ 40 = 47 – x

⇒ x = 7

∴ Height of the platform = 7 m

Given:

The height of the pole is 28 m and it casts a shadow 19.2 m long.

The shadow cast by the tower is 52.5 m long.

Concepts used:

The similarity of triangles using AA(angle-angle) rule

The ratio of corresponding sides of similar triangles is equal.

Calculation:

In ΔABC and ΔEDC,

∠ABC = ∠EDC (each 90°)

∠ACB = ∠ECD (Common angle)

∴ ΔABC ∼ ΔEDC (By AA rule).

⇒ AB/ED = BC/DC

⇒ 52.5/28 = BC/19.2

⇒ BC = 52.5 × 19.2/28

⇒ BC =  36 m.

∴ The length of the shadow cast by another pole of 52.5 m height is 36 m.

Confusion Points

As the position of the sun is fixed at a particular time and both buildings are on the same sides of the sun, the angle made by the shadow with the horizon must be equal.

BE is the height of the bird from the pond.

⇒ BE = ED = h m

⇒ AF = CE = 60 m

⇒ BC = (h – 60) m and CD = (h + 60) m

In ΔABC,

tan 30 = BC/AC

⇒ 1/√3 = (h – 60)/AC

⇒ AC = √3 (h – 60)      ----(i)

In ΔADC

tan 60 = CD/AC

⇒ √3 = (h + 60)/AC

⇒ AC = (h + 60)/√3      ----(ii)

From equation (i) and equation (ii)

√3 (h – 60) = (h + 60)/√3

⇒ 3 (h – 60) = (h + 60)

⇒ 3h – 180 – h = 60

⇒ 2h = 240

⇒ h = 120 m

∴ Height of the bird from the pond is 120 m.

Concept:

If the angle of elevation from a point at distance D from tower of length L is θ 

tan θ = \(\rm L\over D\)

Calculation:

Let the length of the tower OP be x

From the figure,

OA = \(\rm {x\over\tan30}=\sqrt{3}x\)

OB = \(\rm {x\over\tan45}=x\)

OC = \(\rm {x\over\tan60}={x\over√{3}}\)

∴ AB = OA - OB = \(\rm \sqrt 3 x-x\)

BC = OB - OC = \(\rm x-\frac{x}{\sqrt 3}\)

⇒ \(\rm {AB\over BC}={\sqrt 3 x-x\over\rm x-\frac{x}{\sqrt 3}}\)

⇒ \(\rm {AB\over BC}={\sqrt 3(\sqrt 3 -1)\over \sqrt 3 -1} \)

⇒ \(\boldsymbol{\rm AB:BC=\sqrt{3} : 1}\)

Given:

A flagstaff of height h is surmounted on a verticle tower.

Angles of elevation of the bottom and top of the flagstaff are θ and 2θ respectively.

Formula used:

1. tan θ = Perpendicular/Base

2. tan 2θ = \(\frac{(2tan θ)}{(1 - tan^2θ)} \)

3. cos 2θ = \( \frac{1-tan^2θ}{1+tan^2θ}\)

Calculation:

Let the height of the tower be x.

In right triangle ABC

tan θ = AC/AB = x/AB

⇒ AB = x/tan θ          ------(1)

In right triangle ABD

tan 2θ = AD/AB = (x + h)/AB

⇒ \(\frac{(2tan θ)}{(1 - tan^2θ)} \) = (x + h)/AB   [Using formula (2)]

⇒ \(\frac{(2tan θ)}{(1 - tan^2θ)} \) = \(\frac{ (x + h)tan\ θ}{ x} \)     [From equation (1)]

⇒ \(\frac{(2)}{(1 - tan^2θ)} \) = \(1+\frac{ h}{ x} \) 

⇒ \(\frac{h}{x} = \frac{2}{1-tan^2θ}-1\)

⇒  \(\frac{h}{x} = \frac{2-(1-tan^2θ)}{1-tan^2θ}\)

⇒  \(\frac{h}{x} = \frac{1+tan^2θ}{1-tan^2θ}\) 

⇒ \(\frac{x}{h} = \frac{1-tan^2θ}{1+tan^2θ}\) = cos2θ       [Using formula (3)]

⇒ x = hcos 2θ 

∴ The height of the tower is h cos 2θ 

Given:

Height at which kite flying = 138m

Calculation:

Triangle is formed with the height at which the kite flies as the height of the triangle, the Length of string as one side of the triangle, and the horizontal as the base of the triangle.

The triangle is 45 - 45 - 90 triangle.

⇒ sin 45 = (height at which kite fly)/(Length of string)

⇒ 1/√2 = 138m/length of string

⇒ length of sting = 195.16 ≈ 195m

∴ The approximate length of the string is 195m.

Concept : 

\(\rm\ tan\;x =\frac {\text {opposite side }}{\text {adjacent side}}\)

Calculations :

Given the shadow of a tower standing on a level, the plane is found to be 50 m longer when the Sun's elevation is 30° than when it is 60°. 

Suppose, h is the height of the tower and 'l' is the length of the shadow.

from data, we have \(\tan 60° = \frac{h}{L}\)

⇒ \(\rm\sqrt{3} = \frac{h}{L}\)....(1)

similarly, \(\tan 30^\circ = \frac{h}{L + 50}\)

⇒\(\rm\frac{1}{\sqrt{3}}= \frac{h}{L + 50}\) ....(2)

From (1) \(h =\sqrt{3}l\)     -----(3)

Put it in (2)

⇒\(\rm\frac{1}{\sqrt{3}}= \frac{{\sqrt3L}}{L + 50}\)
⇒ L + 50 = 3L

⇒ L = 25 m

⇒ h = 25\(\sqrt{3}\)      [From 3]

∴ The height of the tower is 25\(\sqrt{3}\)

\(\tan 60 = \frac{h}{x}\)

\(\sqrt 3 = \frac{h}{x}\)

\(h = \sqrt 3 x\)

\(\tan 30 = \frac{h}{{x + 40}}\)

\(\frac{1}{{\sqrt 3 }} = \frac{h}{{x + 40}}\)

\(h\sqrt 3 = x + 40\)

\(\left( {\sqrt 3 x} \right)\left( {\sqrt 3 } \right) = x + 40\)

3x = x + 40

2x = 40

x = 20 m