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
- Trigonometric Basics
sinθ=oppositeside/hypotenuse=y/r
cosθ=adjacentside/hypotenuse=x/r
tanθ=oppositeside/adjacentside=y/x
cosecθ=hypotenuse/oppositeside=r/y
secθ=hypotenuse/adjacentside=r/x
cotθ=adjacentside/oppositeside=x/y
From Pythagorean theorem, x2+y2=r2 for the right angled triangle mentioned above
- Basic Trigonometric Values
| θ in degrees | θ in radians | sinθ | cosθ | tanθ |
| 0° | 0 | 0 | 1 | 0 |
| 30° | π/6 | 1/2 | 3/√2 | 1/√3 |
| 45° | π/4 | 1/√2 | 1/√2 | 1 |
| 60° | π/3 | 3/√2 | 1/2 | √3 |
| 90° | π/2 | 1 | 0 | Not defined |
- 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
- 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
- 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
- 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)
- Few Important Values to memorize
√2=1.414, √3=1.732, √5=2.236
Solved Examples
Level 1
| 1.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is: | |
| A. 14.8 m | B. 6.2 m |
| C. 12.4 m | D. 24.8 m |
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
| 2.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is: | |
| A. 346 m | B. 400 m |
| C. 312 m | D. 298 m |
Answer : Option A
Explanation :
tan 30°=RQ/PQ
1/√3=200/PQ
PQ=200√3=200×1.73=346 m
| 3.The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is: | |
| A. None of these | B. 60° |
| C. 45° | D. 30° |
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°
| 4.An observer 2 m tall is 103√ m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is: | |
| A. None of these | B. 12 m |
| C. 14 m | D. 10 m |
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
| 5.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower? | |
| A. 40 m | B. 138.4 m |
| C. 46.24 m | D. 160 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
| 6.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long? | |
| A. 30° | B. 60° |
| C. 45° | D. None of these |
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°
Level 2
| 1.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower? | |
| A. 8 min 17 second | B. 10 min 57 second |
| C. 14 min 34 second | D. 12 min 23 second |
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=800/73minutes ≈10 minutes 57 seconds
| 2.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole? | |
| A. 5 metres | B. 8 metres |
| C. 10 metres | D. 12 metres |
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
| 3.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is: | |
| A. 300 m | B. 173 m |
| C. 273 m | D. 200 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
| 4.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole? | |
| A. 52 m | B. 50 m |
| C. 66.67 m | D. 33.33 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
| 5.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower? | |
| A. 9 m | B. 10.40 m |
| C. 15.57 m | D. 12 m |
Answer : Option A
Explanation :
Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, ABC = 60°, DBC = 30°
Let DC be h.
tan 30°=DC/BC
1/√3=h/BC
h=BC√3—— (1)
tan 60°=AC/BC
√3=(18+h)/BC
18+h=BC×√3—— (2)
(1)/(2)=>h/(18+h)=(BC/√3)/(BC×√3)=1/3
=>3h=18+h=>2h=18=>h=9 m
i.e., the height of the tower = 9 m
| 6.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon? | |
| A. 0.63 meter/sec | B. 2.16 meter/sec |
| C. 3.87 meter/sec | D. 0.72 meter/sec |
Answer : Option B
Explanation :
Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, BCA = 60°
tan 60°=BA/CA
√3=BA/150
BA=150√3
i.e, the distance travelled by the balloon = 150√3meters
time taken = 2 min = 2 × 60 = 120 seconds
Speed = Distance/Time=150√3/120=1.25√3=1.25×1.73=2.16 meter/second
| 7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree? | |
| A. 22 m | B. 44 m |
| C. 33 m | D. None of these |
Answer : Option B
Explanation :
Let DC be the wall, AB be the tree.
Given that DBC = 30°, DAE = 60°, DC = 11 m
tan 30°=DC/BC
1/√3=11/BC
BC = 11√3 m
AE = BC =11√3 m—— (1)
tan 60°=ED/AE
√3=ED/11√3[∵ Substituted the value of AE from (1)]
ED =11√3×√3=11×3=33
Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m
| 8. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles. | |
| A. 141 m and 282 m | B. 70.5 m and 141 m |
| C. 65 m and 130 m | D. 130 m and 260 m |
Answer : Option B
Explanation :
Let AB and CD be the poles with heights h and 2h respectively
Given that distance between the poles, BD = 200 m
Let E be the middle point of BD.
Let AEB = θ and CED = (90-θ) (∵ given that angular elevations are complementary)
Since E is the middle point of BD, we have BE = ED = 100 m
From the right ABE,
tanθ=AB/BE and tanθ=h/100
h = 100tanθ—— (1)
From the right EDC,
tan(90−θ)=CD/ED
cotθ=2h/100[∵tan(90−θ)=cotθ]
2h =100cotθ—— (2)
(1) × (2) => 2h2=1002[∵tanθ×cotθ=tanθ×1/tanθ=1]
=>√2h=100
=>h=100/√2=(100×√2)/( √2×√2)=50√2=50×1.41=70.5
2h=2×70.5=141
i.e., the height of the poles are 70.5 m and 141 m.
| 9. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window? | |
| A. 8.65 m | B. 2 m |
| C. 2.5 m | D. 3.65 m |
Answer : Option D
Explanation :
Let AB be the man and CD be the window
Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°
From the diagram, AF = BE = 5 m
From the right AFD, tan45°=DF/AF
1=DF/5
DF = 5—— (1)From the right AFC, tan60°=CF/AF
√3=CF/5
CF=5√3—— (2)
Length of the window = CD = (CF – DF)
=5√3−5[∵ Substitued the value of CF and DF from (1) and (2)]=5(√3−1)=5(1.73−1)=5×0.73=3.65 m
| 10.The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain? | |
| A. 1.2 km | B. 0.6 km |
| C. 1.4 km | D. 2.7 km |
Answer : Option D
Explanation :
Let A be the foot and C be the summit of a mountain.
Given that CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., CDE = 60°
From the right ABC,
tan45°=CB/AB
=>1=x/AB[∵ CB = x (the height of the mountain)]
=>AB = x—— (eq:1)
From the right AYD,
sin30°=DY/AD
=>1/2=DY/2(∵ Given that AD = 2)
=> DY=1—— (eq:2)
cos30°=AY/AD=>√3/2=AY/2(∵ Given that AD = 2)=> AY=√3—— (eq:3)
From the right CED, tan60°=CE/DE=>tan60°=(CB – EB)/YB∵ [CE = (CB – EB) and DE = YB)]
=>tan60°=(CB – DY)/(AB – AY)[ ∵ EB = DY and YB = (AB – AY)]
=>tan60°=(x – 1)/(x -√3)∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = 3√(eq:3)]
=>√3=(x – 1)/(x -√3)=>x√3−3=x−1=>x(√3−1)=2=>0.73x=2=>x=2/0.73=2.7
i.e., the height of the mountain = 2.7 km
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