Height and Distance solved problems - WTSkills- Learn Maths, Quantitative Aptitude, Logical Reasoning

I this post we will discuss quantitative aptitude questions of height and distance. All the questions below are solve step by step with diagram illustration for your better understanding.

As the name suggests, the height and distance chapter basically involves calculating height, distance and angle between different objects. The questions can be solved easily by taking help of trigonometry functions such as sin, cos or tan function.
Before solving the chapter, i would strongly suggest to brush up your concepts of trigonometry as they would be widely used in the below questions.

Height and Distance Quantitative Aptitude

Q1. A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time, a tower casts a shadow 40 m long on the ground. The height of the tower is

(a) 72 m (b) 60 m

height and distance quantitative aptitude questions for competition exam like GRE, GMAT, CAT, MAT, CMAT, NMAT, SNAP ,SSC, SSC-CGL, SSC-CHSL, IBPS, Bank exam, SBI PO, SBI clerk, AFCAT, NDA, NABARD

In\triangle ACB\quad and\quad \triangle PCQ\\\ \\ \angle C=\angle C\quad (common)\\ \\ \angle ABC=\angle PQC\quad (right\quad angle)\\ \\ \triangle ACB\sim \triangle PCQ\quad (AA\quad similarity)\\\ \\ \frac { AB }{ PQ } =\frac { BC }{ QC } \\\ \\ \frac { h }{ 12 } =\frac { 4000 }{ 8 } \\\ \\ h=\frac { 4000\times 12 }{ 8 } \\\ \\ h=6000\quad cm\quad =60m

option (b) is the right answer

Q2 The angle of elevation of tower from a distance 50 m from its foot is 30º. Then the height of the tower is

(a) 50 m (b) 60 m

height and distance solved aptitude questions

Let\quad AB\quad the\quad height\quad of\quad the\quad tower\\ \\ BC=\quad 50m\\\ \\ tan30=\frac { AB }{ BC } \\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ 50 } \\ \\ AB=\frac { 50 }{ \sqrt { 3 } } \quad m

option (d) is right answer

Q3. The length of the shadow of a vertical tower on level ground increases by 10 meters when the altitude of the sun changes from 45º to 30º. Then the height of the tower is

(a) 5 m (b) $5(\sqrt { 3 } -1)\quad m$

how to solve height and distance problems

CD=\quad h\\\ \\ In\quad right\quad angle\quad \triangle BCD\\ \\ tan45=\frac { DC }{ BC } \\ \\ \frac { h }{ x } =1\\ \\ h=x\quad \quad ….(a)\\\ \\ In\quad right\quad angle\quad \triangle ACD\\ \\ tan30=\frac { DC }{ AC } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\quad \frac { h }{ x+10 } \\\ \\ \sqrt { 3h } =\quad h+10\quad \quad \quad \quad [h=x]\\\ \\ h=\frac { 10 }{ \sqrt { 3 } -1 } \\\ \\ h=\frac { 10 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \\\ \\ h=\frac { 10\sqrt { 3 } +1 }{ 3-1 } \\\ \\ h=5(\sqrt { 3 } +1)\quad m

option (c) is right answer

Q4. The top of two poles of height 24m and 36m are connected by a wire. If the wire makes an angle of 60º with the horizontal, then the length of the wire is

(a) 6m (b) 8 * Square root 3

height and distance questions for competition exams

AB = 24 meter
CD = 36 meter
BD = length of wire

DE = CD – CE
DE = CD – AB [CE= AB]
DE = 36 – 24
DE = 12 meter

In\triangle BDE\\\ \\ sin60=\frac { DE }{ BD } \\ \\ \frac { \sqrt { 3 } }{ 2 } =\frac { 12 }{ BD } \\\ \\ BD=\frac { 12\times 2 }{ \sqrt { 3 } } \\ \\ BD=4\sqrt { 3 } \times 2=8\sqrt { 3 } m\\\ \\ The\quad length\quad of\quad the\quad wire\quad is\quad 8\sqrt { 3 } m

option (b) is the right answer

Q5. From the top of a hill 200m high the angle of depression of the top and the bottom of a tower are observed to be 30º and 60º. The height of the tower is (in m)

(a) $133\frac { 1 }{ 3 } \quad m$ (b) 166

height and distance question is important application of trigonometry chapter

Solution
AB = 200 meter (hill)
Angle ADE = 30
Angle ACB = 60
DE= BC = x meter

In\triangle ABC\\ \\ tan60=\frac { AB }{ BC } \\ \\ \sqrt { 3 } =\frac { 200 }{ x } \\ \\ x=\frac { 200 }{ \sqrt { 3 } } \quad m\\\ \\ In\triangle AED\ tan30=\frac { AE }{ DE } \\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { AE }{ ({ 200 }/{ \sqrt { 3 } }) } \\\ \\ AE=\frac { { 200 }/{ \sqrt { 3 } } }{ \sqrt { 3 } } \\ \\ AE=\frac { 200 }{ 3 } \quad m\ \ \\CD=\quad AB-AE\\ \\ CD=200-\frac { 200 }{ 3 } =\frac { 400 }{ 3 } =133\frac { 1 }{ 3 } \quad m\

Q6. From a point 20m away from the foot of a tower, the angle of elevation of the top of the tower is 30º. The height of the tower is

(a) 10 m (b) 20 m

Read Solution

Let AB be the height of tower
BC = 20 meter

In right triangle ABC
$tan30=\frac { AB }{ BC } \\\ \\ \sqrt { 3 } =\frac { AB }{ 20 } \\\ \\ AB=20\sqrt { 3 } \quad m$

option (d) is the right answer

Q7. The angle of elevation of ladder leaning against a house is 60º and the foot of the ladder is 6.5 meters from the house. The length of the ladder is

(a) 11 m (b) 13 m

(d) 15 m (d) 3.25 m

Read Solution

quantitative aptitude questions for competition exams like GMAT, GRE, Math Olympiad, SSC, SSC-CGL, SSC-CHSL, NMAT, SNAP, CAT, CMAT, NABARD, AFCAT, NDA

AC = Ladder
BC = 6.5 meter

In triangle ABC
$cos60=\frac { BC }{ AC } \\\ \\ \frac { 1 }{ 2 } =\frac { 6.5 }{ AC } \\\ \\ AC=6.5\times 2\\\ \\ AC=13\quad m$

option (b) is the right answer

Q8. The length of the shadow of a vertical tower on level ground increases by 10 meters when the altitude of the sun changes from 45º to 30º. Then the height of the tower is

(a) $5\sqrt { 3 } +1\quad m$ (b) $5\sqrt { 3 } -1\quad m$

Read Solution

AB=h\\\ \\ In\triangle ABC\\ \\ tan45=\frac { AB }{ BC } \\\ \\ 1=\frac { AB }{ BC } \\\ \\ AB:BC=1:1\quad \quad \quad ….(a)\\\ \\ In\triangle ABD\\ \\ tan30=\frac { AB }{ BD } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ BD } \\\ \\ AB:BD=1:\sqrt { 3 } \quad \quad \quad ….(b)\\\ \\ \ BC\qquad :\qquad AB\qquad :\qquad BD\\ \\ \quad 1\qquad :\qquad \quad 1\qquad :\qquad \quad 1\\ \\ 1\qquad \quad :\quad \qquad 1\qquad :\qquad \sqrt { 3 } \\ \\ 1\qquad :\qquad 1\qquad :\qquad \quad \sqrt { 3 } \\\ \\ AB:BC:BD=x:x:\sqrt { 3 } x\\\ \\ CD=BD-BC\\\ \\ CD=\sqrt { 3 } x-x\\\ \\ (\sqrt { 3 } -1)x\quad =10\\\ \\ x=\frac { 10 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \\\ \\ x=\frac { 10\sqrt { 3 } +1 }{ 3-1 } =5\sqrt { 3 } +1\quad m

Q9. If a pole of 12 m height cast a shadow of 4m long on the ground then the sun’s angle of elevation at that instant is

(a) 30º (b) 60º

In\triangle ABC\\\ \\ tan\theta =\frac { AB }{ BC } \\\ \\ tan\theta =\frac { 12 }{ 4\sqrt { 3 } } \\\ \\ \ tan\theta =\sqrt { 3 } \\\ \\ tan\theta =tan60\\\ \\ \theta =60

option (b) is the right answer

Q10. From the top of a tower of height 180 m the angle of depression of two objects on either sides of the tower are 30º and 45º. Then the distance between the objects are

(a) $180(\sqrt { 3 } +1)m$ (b) $180(\sqrt { 3 } -1)m$

Read Solution

height and distance chapter of mathematics

AB=180\\\ \\ In\triangle ABC\\ \\ tan45=\frac { AB }{ BC } \\ \\ 1=\frac { AB }{ BC } \\ \\ AB:BC=1:1\quad \quad \quad ….(a)\\\ \\ \\\ \\ In\triangle ABD\\ \\ tan30=\frac { AB }{ BD }\\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ BD } \\\ \\ AB:BD=1:\sqrt { 3 } \quad \quad \quad ….(b)\\\ \\ \\\ \\ \ BC\qquad :\qquad AB\qquad :\qquad BD\\ \\ \quad 1\qquad :\qquad \quad 1\qquad :\qquad \quad 1\\ \\ 1\qquad \quad :\quad \qquad 1\qquad :\qquad \sqrt { 3 } \\ \\ 1\qquad :\qquad 1\qquad :\qquad \quad \sqrt { 3 } \\\ \\ AB:BC:BD=x:x:\sqrt { 3 } x\\\ \\ AB=x=180\\\ \\ CD=BD+BC\\ \\ CD=\sqrt { 3 } x+x\\ \\ CD=(\sqrt { 3 } +1)x\\ \\ CD=180(\sqrt { 3 } +1)m

Height and Distance Quantitative Aptitude

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