Height and Distance, Quantitative Aptitude, Civil Service Examination, RPSC - UPSC PDF Download

This topic has many practical application in day to day life. In engineering stage it is used in surveying. The basic purpose is to find the unknown variables by observing the angle of the line of sight. This is done by using some the fact that in a right angled triangle the ratio of any two sides is a function of the angle between them. From exam point of view this is one of the more tough sections and tedious to some extent. So an aspirant must thoroughly solve all the questions given here.

Important Formulas

1. Trigonometric Basics

sinθ = opposite side/hypotenuse=y/r

cosθ = adjacent side/hypotenuse=x/r

tanθ = opposite side/adjacent side=y/x

cosecθ = hypotenuse/opposite side=r/y

secθ = hypotenuse/adjacent side=r/x

cotθ = adjacent side/opposite side=x/y

From Pythagorean theorem, x²+y²=r² for the right angled triangle mentioned above

2. Basic Trigonometric Values

3. Trigonometric Formulas

Degrees to Radians and vice versa

360°=2π radian

Trigonometry - Quotient Formulas

tanθ=sinθ/cosθ

cotθ=cosθ/sinθ

Trigonometry - Reciprocal Formulas

cosecθ=1/sinθ

secθ=1/cosθ

cotθ=1/tanθ

Trigonometry - Pythagorean Formulas

sin²θ+cos²θ=1

sec²θ−tan²θ=1

cosec²θ−cot²θ=1

4. Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer's eye (the line of sight).

i.e., angle of elevation =  ∠AOP

5. Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer's line of sight

i.e., angle of depression = ∠AOP

6. Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then BD/DC=AB/AC

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e.,  BAD =  CAD in the above diagram)

7. Few Important Values to memorize

√2=1.414, √3=1.732, √5=2.236

Solved Examples
 

Answer : Option D

Explanation :

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQ/PR

1/2=12.4/PR

PR=2×12.4=24.8 m
 

Answer : Option A

Explanation :

tan 30°=RQ/PQ

1/√3=200/PQ

PQ=200√3=200×1.73=346 m
 

Answer : Option C

Explanation :

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ

Let  ∠QPR = θ

tan θ = QR/PQ=1 (since QR = PQ)

=> θ = 45°

i.e., required angle of elevation = 45°
 

Answer : Option B

Explanation :

SR = PQ = 2 m

PS = QR = 10√3m

tan 30°=TS/PS

1/3=TS/10√3

TS=10√3/√3=10 m

TR = TS + SR = 10 + 2 = 12 m
 

Answer : Option B

Explanation :

Let AC be the tower and B be the position of the bus.

Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, ∠DAB = 30° 

∠ABC =  ∠DAB = 30° (Because DA || BC)

tan 30°=AC/BC =>tan 30°=80/BC =>BC = 80/tan 30°=80/(1/√3)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m
 

Answer : Option B

Explanation :

Let RQ be the pole and PQ be the shadow 

Given that RQ = 18 m and PQ = 6√3 m

Let the angle of elevation,  RPQ = θ 

From the right   PQR, 

tanθ=RQ/PQ=18/6√3=3/√3=(3×√3)/(√3×√3)=3√3/3=√3

θ=tan−1(3√)=60°
 

Answer : Option B

Explanation :

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car

Then, ∠ADC = 30° , ∠ACB = 45°

Let AB = h, BC = x, CD = y

tan 45°=AB/BC=h/x

⇒1= h/x ⇒ h=x------(1)

tan 30°=AB/BD = AB/(BC + CD)=h/(x+y)

⇒1/√3 = h/(x+y)

⇒x + y = √3h

⇒y = √3h – x

⇒ y = √3h−h(∵ Substituted the value of x from equation 1 )

⇒ y = h(√3−1)

Given that distance y is covered in 8 minutes i.e, distance h(√3−1) is covered in 8 minutes

Time to travel distance x = Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8---(A)

h∝t---(B)

(A)/(B) ⇒ h(√3−1)/h = 8/t

⇒ (√3−1) = 8/t

⇒ t = 8/(√3−1) = 8/(1.73−1) = 8/.73 = 800/73 minutes ≈10 minutes 57 seconds
 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the tower and DE represents the pole

Given that AC = 15 m , ∠ADB = 30°, ∠AEC = 60°

Let DE = h

Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE

tan 60°=AC/CE ⇒√3=15/CE ⇒CE = 15√3--- (1)

tan 30°=AB/BD ⇒1/√3=(15−h)/BD

⇒1/√3=(15−h)/(15/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

⇒(15−h)=(1/√3)×(15/√3)=15/3=5

⇒h=15−5=10 m

i.e., height of the electric pole = 10 m
 

Answer : Option C

Explanation :

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, ∠BAD = 30° , ∠BCD = 45°

tan 30° = BD/BA ⇒1/√3=100/BA

⇒ BA=100√3

tan 45° = BD/BC

⇒1=100/BC

⇒ BC=100

Distance between the two ships = AC = BA + BC 
=100√3+100=100(√3+1)
= 100(1.73+1)
= 100×2.73=273 m
 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m 

XAD =  ADB = 30° (∵ AX || BD ) 
XAE =  AEC = 60° (∵ AX || CE) 

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE

⇒ √3=100/CE ⇒ CE = 100/√3--- (1)

tan 30°=AB/BD ⇒ 1/√3=(100−h)/BD

⇒1/√3=(100−h)/(100/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

⇒(100−h)=1/√3×100/√3=100/3=33.33⇒h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m