Height and Distance Questions, Formulas and Shortcut Tricks

Height and Distance Formulas

One of the important applications of trigonometry is in finding the height and distances of the point which are not directly measurable. This is done with the help of trigonometric ratios.


(A) Angle of Elevation
O: Position of the object
A: Position of the height
Here, object (A) is at height level than observer (O)
OX: Reference line (or horizontal line)
OA: Line of sight (or line of observation)
\theta= \angleAOX =angle of elevation

Angle of Depression

O: Position of the observer
B: Position of the object
Here, object (B) is at lower level than the observer (O)
OX: Reference line (or horizontal line)
OB: Line of sight (or line of observation)
\beta = \angleBOX =angle of depression


(A) In a right-angle triangle ABC,*


(B) In any triangle ABC,

 \frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C} [ Sin Rule ]

 \Rightarrow \:\frac{Length\:\:of\:\:any\:\:side}{Sine\:\:of\:\:the\:\:angle\:\:opposite\:\:to\:\:the\:\:side} = constant (of each side of the angle)

(C) In any angle ABC

If \frac{AD}{DC}=\frac{m}{n} and
\angleBAD =\alpha
\angleCAD =\beta
\angleADC =\theta
Then, (m + n) cot \theta=m cot\alpha-n cot\beta

(D) In a right - angle triangle ABC,

If DE \left |\right | AB, then



Important Examples of Height and Distance

Example 1: A ladder 15 m long just reaches the top of a wall and makes an angle of 60^{\circ} with the wall. Find the distance of the foot of the ladder from the wall. (\sqrt{3}=1.732)

Solution: Let AB be the wall

AC = length of ladder = 15m
 \angleBAC = 60 ^{\circ}

BC = x = distance of the foot of the ladder from the wall
Then, using the relation,
 sin\:\theta =\frac{p}{h}=\frac{BC}{AC}
x = 15 sin\:\:60^{\circ} = 15\times\frac{\sqrt{3}}{2}=15

Example 2: From a tower 125 m high, the angle of depression of a car is 30^{\circ}. Find how far the car is from the tower.

Solution: Let OT be the tower and C be the position of the car

\angle XTC = angle of depression of car (viewed from T )

=30^{\circ} (given)
= \angle TCO (since XT\left |\right| CO,)
OT = length of tower =125 m
OC = x = distance of the car from the fool of the tower OT

Then, using the relation

(from time saving, write directly as other leg (x) =\frac{125}{tan\:30^{\circ}} Refer aid to momory of 25.2)


\Rightarrow\:\:x=\frac{125\times \sqrt{3}}{1} = 216.5
Hence, the required distance is 216.5 m

Example 3: The angle of elevation of the top of a tower 30 m high, from two points on the level ground on its opposite side are 45 ^{\circ} and 60 ^{\circ}. Find the distance between the two point

Let OT be the tower

 \therefore Height of tower = OT = 30 m

Lets A and B be the opposite side of tower OT.
Then, angle of elevation from A =  \angleTAO =45 ^{\circ}
and angle of elevation from B =  \angleTBO =60 ^{\circ}
Distance between AB = AO + OB = x + y (say)

Now, in right \DeltaATO,

Tan = 45 ^{\circ} = \frac{OT}{AO} = \frac{30}{x}
\Rightarrow x =\frac{30}{tan\: 45 ^{\circ}} = 30 m

And in right \DeltaBTO,

 tan\: 60 ^{\circ} = \frac{OT}{AO} = \frac{30}{y}
\Rightarrow y
=\frac{30}{ tan\: 60 ^{\circ} } = \frac{30}{\sqrt{3}}=\frac{30\sqrt{3}}{3}=17.32 m

Hence the required distance = x + y = 30 + 17.32 = 47.32 m

Example 4: A balloon leaves the earth at point A and rise at a uniform velocity. At the end of  1\frac{1}{2}min. An observer situated at a distance of 200 m from A, finds the angular elevation of the balloon to the 60^{\circ}.Find the speed of the balloon.

Let the balloon, after leaving the point A, reach to point B vertically upward in  1\frac{1}{2}min. The observer situated at point O, observes the angle of elevation of B as 60^{\circ}

\angle BOA =60^{\circ}

tan60^{\circ} =\frac{AB}{OB}

AB = OBtan60^{\circ}=200 \times \sqrt{3} metre Speed of the balloon

=\frac{distance}{time} =\frac{AB}{time to coss AB}

=\frac{200\times \sqrt{3}}{\frac{3}{2}\times 60} metre/sec

=3.87 metre/sec.


Height and Distance Questions from Previous Year Exams

  • Height and Distance Aptitude


Height and Distance Important Video


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