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

In this post we will discuss the questions of height and distance which is part of the syllabus of quantitative aptitude. All the questions given below are important as they are previously asked in competition exams like GRE, GMAT, Math Olympiad, CAT, MAT, CMAT, SNAP, NMAT, SSC, SSC-CGL, SSC-CHSL, Banking , IBPS, SBI PO, SBI Clerk, RRB, NDA, AFCAT etc.

The problems of height and distance require knowledge of basic trigonometry function. If you are unaware of the concept of trigonometry, i would strongly suggest to complete that chapter first and then move on with height and distance questions

Height & Distance Aptitude Questions

Q1. One flies a kite with a thread 150 meter long. If the thread of kite makes an angle of 60º with the horizontal line, then the height of the kite from the ground (assuming the thread to be in a straight line) is

(a) 50 meter (b) $75\sqrt { 3 }$ meter

Read Solution

AB=150,\quad \angle BAC=60\\\ \\ In\triangle ABC\\\ \\ sin60=\frac { BC }{ AB } \\\ \\ \frac { \sqrt { 3 } }{ 2 } =\frac { BC }{ 150 } \\\ \\ BC=150\times \frac { \sqrt { 3 } }{ 2 } \\\ \\ BC=75\sqrt { 3 } meter

option (b) is the right answer

Q2. The angle of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are respectively 15º and 30º. If A and B are on the same side of the tower and AB= 48 meter, then the height if the tower is;

(a) 25 meter (b) 27 meter

how to solve height and distance questions

PQ=h\\ \\ QB=x\\\ \\ In\triangle APQ,\\\ \\ tan15=\frac { PQ }{ AQ } =\frac { h }{ x+48 } \quad \quad \quad \quad [tan15=tan(45-30)=2-\sqrt { 3 } ]\\\ \\ 2-\sqrt { 3 } =\frac { h }{ x+48 } \quad \quad …..(a)\\\ \\ \quad \ In\triangle PQB,\\ \\ tan30=\frac { PQ }{ QB } =\frac { h }{ x } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x } \\\ \\ \sqrt { 3 } h=x\quad …..(b)\\\ \\ 2-\sqrt { 3 } =\frac { h }{ \sqrt { 3 } h+48 } \\\ \\ 2-\sqrt { 3 } (\sqrt { 3 } h+48)=h\\\ \\ \ h=\frac { 48 }{ 2 } =24meter

option (c) is the right answer

Q3. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º. The man walks some distance towards the tower and then his angle of elevation of the top of the tower is 60º. If the height of the tower is 30m, then the distance he moves is?

(a) 22 m (b) 22 m

trigonometry and height and distance application questions

h=30m\\\ \\ In\quad \triangle ABP\\ \\ tan30=\frac { AB }{ BP } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ BP } \\\ \\ AB:BP=1:\sqrt { 3 } \quad ……(a)\\\ \\ \ In\quad \triangle ABC\\ \\ tan60=\frac { AB }{ BC } \\\ \\ \sqrt { 3 } =\frac { AB }{ BC } \\\ \\ AB:BC=\sqrt { 3 } :1\quad …..(b)\\\ \\ \ BP\qquad :\qquad AB\qquad :\qquad BC\\ \\ \sqrt { 3 } \quad \quad :\qquad 1\qquad \quad :\quad \quad \quad 1\\\ \\ \sqrt { 3 } \quad \quad :\quad \sqrt { 3 } \qquad :\qquad 1\\\ \\ \quad 3\qquad :\quad \sqrt { 3 } \qquad :\qquad 1\\\ \\ AB=\sqrt { 3 } x\quad =\quad 30meter\\\ \\ x=\frac { 30 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =10\sqrt { 3 } \\\ \\ PC=BP-PC\\ \\ PC=3x-x\\ \\ PC=2x\\\ \\ 2x=10\sqrt { 3 } \times 2=20\sqrt { 3 } meter

option (d) is the right answer

Q4. An Aeroplane when flying at a height of 3125m from the ground passes vertically below another plane at an instant when the angle of elevation of two planes from the same point on the ground is 30º and 60º respectively. The distance between the two planes at that instant is?

(a) 6520m (b) 6000m

Height and distance aptitude questions for competition exams like GRE, GMAT, Math Olympiad, CAT, MAT, CMAT, NMAT, SNAP. SSC, SSC-CGL, SSC-CHSL, Bank exams, IBPS, SBI PO, SBI clerk, RRB, NABARD, AFCAT, NDA exams

BC=3125m\\\ \\ let\quad AC=x\\\ \\ In\quad \triangle ABD\\ \\ tan60=\frac { AB }{ BD } \\ \\ \sqrt { 3 } =\frac { 3125+x }{ BD } \\\ \\ BD=\frac { 3125+x }{ \sqrt { 3 } } \\\ \\ \ In\quad \triangle BCD\\ \\ tan30=\frac { BC }{ BD } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { 3125 }{ \frac { 3125+x }{ \sqrt { 3 } } } \\\ \\ \ 3125+x=3\times 3125\\\ \\ \ \ x=6250m\

option (d) is the right answer

Q5. The shadow of the tower becomes 60 meters longer when the altitude of the sun changes from 45º to 30º. Then the height of the tower is?

Read Solution

\ \ AB=h\\\ \\ \ In\triangle ABC\\\ \\ tan45=\frac { AB }{ BC } \\ \\ 1=\frac { AB }{ BC }\\\ \\ \quad \ AB=BC\quad or\quad AB:BC=1:1\quad …..(a)\\\ \\ In\triangle ABD\\ \\ tan30=\frac { AB }{ BD } \\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ BD } \quad \quad \quad \\ \\ AB:BD=1:\sqrt { 3 } \quad ……(b)\\\ \\ BD\qquad :\qquad AB\qquad :\qquad BC\\ \\ \quad 1\qquad :\qquad \quad 1\qquad :\qquad \quad 1\\ \\ \sqrt { 3 } \quad :\qquad 1\qquad \quad :\qquad \quad 1\\ \\ \sqrt { 3 } :\qquad 1\qquad \quad :\qquad 1\\\ \\ CD=BD-BC\\ \\ CD=\sqrt { 3 } x-x\\ \\ (\sqrt { 3 } -1)x=60 \quad \quad \quad \quad [h=x,\quad AB=BC]\\ \\ h=\frac { 60 }{ \sqrt { 3 } -1 } \\\ \\ h=\frac { 60(\sqrt { 3 } +1) }{ (\sqrt { 3 } -1)(\sqrt { 3 } +1) } \\\ \\ \ h=30(\sqrt { 3 } +1)\quad meter\ \ \ \ \ \

Q6. The shadow of a tower is $\sqrt { 3 }$ times its height. Then the angle of elevation of the top of the tower is

(a) 45º (b) 30º

AB=x\\ \\ BC=\sqrt { 3 } x\\\ \\ tan(ACB)=\frac { AB }{ BC } \\ \\ tan(ACB)=\frac { x }{ \sqrt { 3 } x } =\frac { 1 }{ \sqrt { 3 } } \\\ \\ tan(ACB)=tan30\\\ \\ \angle ACB=30

option (b) is the right answer

Q7. The angle of elevation of an aeroplane from a point on the ground is 60º. After 15 seconds flight, the elevation changes to 30º, if the aeroplane is flying at a height of 1500 m, find the speed of the plane

(a) 300 m/sec (b) 200 m/sec

Height and distance aptitude questions and answer

BD=CE=1500\sqrt { 3 } \quad (height)\\ \\ \angle BAE=60,\quad \angle CAE=30\\\ \\ \ In\triangle ADB\\ \\ tan60=\frac { BD }{ AD } =\frac { 1500\sqrt { 3 } }{ AD } \\\ \\ \sqrt { 3 } =\frac { 1500\sqrt { 3 } }{ AD } \\\ \\ AD=\frac { 1500\sqrt { 3 } }{ \sqrt { 3 } } =1500\quad m\\\ \\ \\\ \\ In\triangle \quad CAE\\ \\ tan30=\frac { CE }{ AE } =\frac { 1500\sqrt { 3 } }{ AE } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { 1500\sqrt { 3 } }{ AE } \\ \\ \ AE=1500\times 3=4500\quad m\\\ \\ Distance\quad covered\quad by\quad plane\quad in\quad 15\quad seconds;\\\ \\ \quad \ BC=AE-AD=4500-1500=3000m\ Speed\quad of\quad aeroplane\quad =\frac { 3000 }{ 15 } =200\quad m/sec

Q8. There are two temples, one on each bank of a river just opposite to each other. One temple is 54m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30º and 60º respectively. The length of the temple is;

(a) 18m (b) 36 m

Height and distance questions, formulas and shortcut tricks

AB = 54 m
CD = BE = h
BC = DE = x

tan60=\frac { AB }{ BC } \\ \\ \sqrt { 3 } =\frac { 54 }{ x } \\ \\ x=\frac { 54 }{ \sqrt { 3 } } =18\sqrt { 3 } m\\\ \\ In\triangle ADE\\ \\ tan30=\frac { AE }{ DE } \\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { 54-h }{ 18\sqrt { 3 } }\\ \\ 54-h=18\\ \\ h=54-18=36m

Q9. The angle of elevation of a tower from a distance 100m from its foot is 30º. Height of the tower is

(a) 300 m (b) 50 m

application of trigonometry height and distance questions

Let\quad h\quad be\quad the\quad height\quad of\quad the\quad tower.\\\ \\ tan60=\frac { AB }{ BC } \\ \\ \sqrt { 3 } =\frac { h }{ 100 } \\ \\ h=100\sqrt { 3 } m

Height and Distance – Aptitude Questions

Height & Distance Aptitude Questions

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Height & Distance Aptitude Questions

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