From this topic, usually, questions from heights and distances is asked. Sometimes, it has been observed that straight forward questions from trigonometric ratios such as questions based on quadrants, small simplification questions, etc. have also been asked in addition to heights and distances. Concepts of trigonometry are applicable if and only if the triangle is a right angled triangle.
- π radian = 180 degrees
- sin θ = Perpendicular / Hypotenuse
cos θ = Base / Hypotenuse
tan θ = Perpendicular / Base
- In the first quadrant, all trigonometric ratios (sin, cos, tan, cosec, sec, cot) are positive.
In second quadrant, only sin and cosec are positive.
In third quadrant, only tan and cot are positive.
In fourth quadrant, only cos and sec are positive.
- sin2θ + cos2θ = 1
1 + tan2θ = sec2θ
1 + cot2θ = cosec2θ
- sin(- θ) = – sin θ
cos(- θ) = cos θ
tan(- θ) = – tan θ
cosec(- θ) = – cosec θ
sec(- θ) = sec θ
cot(- θ) = – cot θ
- Here, θ1 is called the angle of elevation and θ2 is called the angle of depression.
- For one specific type of problem in height and distances, we have a generalized formula.
Height = Distance moved / [cot(original angle) – cot(final angle)]
=> h = d / (cot θ1 – cot θ2)
Example : A man was standing at a point 100 m away from the building. From that point, the angle of elevation of the top of the building was 30 degrees. On moving 30 m towards the building, the angle of elevation changed to 45 degrees. Find the height of the building.
Solution : Height = 30 / (cot 30 – cot 45) = 30 / ( – 1) = 15 + 15 m
- Trigonometry & Height and Distances | Set-2
- LCM and HCF
- Work and Wages
- Pipes and Cisterns
- Trains, Boats and Streams
- Ratio Proportion and Partnership
- Mixture and Alligation
- Profit and Loss
- Permutation and Combination
- Assumptions and Conclusions, Courses of Action
- Error Detection and Correction
- Permutation and Combination | Set-2
- Cvent Interview Experience (On campus for Internship and Full Time)
- TCS Ninja Interview Experience and Interview Questions
- Problem on HCF and LCM
- Problem on Pipes and Cisterns
- Problem on Time Speed and Distance
- Problem on Trains, Boat and streams
- Ratio proportion and partnership | Set-2
- Mixture and Alligation | Set 2
Heights and Distances
Problems on height and distances are simply word problems that use trigonometry.
Question 1 : Simplify : [ (cos 80) / (sin 10) ] + cos 59 cosec 31
Solution : [ (cos 80) / (sin 10) ] + cos 59 cosec 31 = [ (cos (90 – 10)) / (sin 10) ] + cos 59 cosec (90 – 59)
=> [ (cos 80) / (sin 10) ] + cos 59 cosec 31 = (sin 10 / sin 10) + cos 59 sec 59
=> [ (cos 80) / (sin 10) ] + cos 59 cosec 31 = (sin 10 / sin 10) + cos 59 (1 / cos 59)
=> (sin 10 / sin 10) + cos 59 sec 59 = 1 + 1 = 2
Question 2 : From the top of alight house, the angles of depression of two ships are 30 and 45 degrees. The two ships, as it was observed from the top of the light house, were 100 m apart. Find the height of the light house.
Solution : here, we can apply the formula Height = Distance / [cot(original angle) – cot(final angle)]
=> Height of the light house = 100 / (cot 30 – cot 45) = 100 / ( – 1) = 50 + 50 m
Question 3 : A 80 m long ladder is leaning on a wall. If the ladder makes an angle of 45 degrees with the ground, find the distance of the ladder from the wall.
Here, cos θ = Base / Hypotenuse
=> cos 45 = Base / 80
=> Base = 80 cos 45 = 80 / = 40
=> Distance of the ladder from the wall = 40 m
Question 4 : There are two poles, one on each side of the road. The higher pole is 54 m high. From the top of this pole, the angle of depression of top and bottom of the shorter pole is 30 and 60 degree respectively. Find the height of the shorter pole.
Let AB and CD be the two poles.
Let AC = x m and CD = h m
Now, in triangle ABC,
tan 60 = AB / AC
=> = 54 / AC
=> AC = 18 m
Clearly, AC = DE = 18 m
In triangle BED,
tan 30 = BE / DE
=> BE = DE tan 30
=> BE = 18 / m
=> BE = 18 m
=> CD = AE = AB – BE
=> CD = 54 – 18 = 36 m
Therefore, height of the shorter pole = 36 m
This article has been contributed by Nishant Arora
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