Skip to content
Top 100 Most Asked QuestionsLaunching soon20% off the yearly Test Series pass
Klay
Start Learning

LearnQuantitative AptitudeGeometry & Mensuration

04.11 · Geometry & Mensuration

Heights & Distances

Elite lesson11 formulas10 questions

On the syllabus of

Elite lesson56 minHinglish

Formulas

At a glance 11 formulas

  1. 1Height h, distance dh=dtan⁡θ, θ=elevationh = d\tan\theta,\ \theta = \text{elevation}d=hcot⁡θ, also the shadowd = h\cot\theta,\ \text{also the shadow}
  2. 2Depression equals elevationThe angle of depression from A to B equals the angle of elevation from B to A (alternate angles), so a look down uses the same right triangle.
  3. 3Handy angle factsAt 45∘45^\circ the height equals the distance. Since tan⁡60∘=3tan⁡30∘\tan 60^\circ = 3\tan 30^\circ, an object is 3 times as far away when seen at 30∘30^\circ as at 60∘60^\circ.
  4. 4Walking towards ith=dcot⁡α−cot⁡βh = \dfrac{d}{\cot\alpha - \cot\beta}d=walk in; elevation α rises to βd = \text{walk in};\ \text{elevation } \alpha \text{ rises to } \beta
  5. 5Opposite sidesh=dcot⁡α+cot⁡βh = \dfrac{d}{\cot\alpha + \cot\beta}d=gap; elevations α,β from either sided = \text{gap};\ \text{elevations } \alpha, \beta \text{ from either side}
  6. 6Complementary elevationsh=xyh = \sqrt{xy}elevations θ, 90∘−θ at distances x,y\text{elevations } \theta,\ 90^\circ - \theta \text{ at distances } x, y
  7. 7Split at eye levelH=h(1+tan⁡αtan⁡β)H = h\left(1 + \dfrac{\tan\alpha}{\tan\beta}\right)H=tower, h=eye level; top α, foot βH = \text{tower},\ h = \text{eye level};\ \text{top } \alpha,\ \text{foot } \beta
  8. 8Pole on a roofpole=d(tan⁡β−tan⁡α)\text{pole} = d(\tan\beta - \tan\alpha)d=distance; roof at α, tip at βd = \text{distance};\ \text{roof at } \alpha,\ \text{tip at } \beta
  9. 9Steady approachFor an object moving straight at the foot at a steady speed, the distance left is hcot⁡θh\cot\theta, so times are in the ratio of the drops in cot⁡θ\cot\theta; hh cancels.
  10. 10Cloud and its reflectionH=h tan⁡β+tan⁡αtan⁡β−tan⁡αH = h\,\dfrac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}H=cloud, h=eye; cloud α, image βH = \text{cloud},\ h = \text{eye};\ \text{cloud } \alpha,\ \text{image } \beta
  11. 11South and eastPoints due south and due east of a tower are hcot⁡αh\cot\alpha and hcot⁡βh\cot\beta from its foot, so they are hcot⁡2α+cot⁡2βh\sqrt{\cot^2\alpha + \cot^2\beta} apart (Pythagoras on the ground).

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    From a point on level ground 40 m from the foot of a temple tower, the angle of elevation of its top is 60∘60^\circ. How tall is the tower?

    Correct answer40340\sqrt{3} m

    Why

    Height is level distance times the tangent of the elevation.

    h=40×tan⁡60∘=40×3h = 40 \times \tan 60^\circ = 40 \times \sqrt{3}; using tan⁡30∘\tan 30^\circ by mistake gives the trap 403\frac{40}{\sqrt{3}}.

    The tower is 403≈69.340\sqrt{3} \approx 69.3 m tall.

  2. Q2Easy

    Asked in IPMAT Indore 2024

    The angle of elevation of the top of a pole from a point A on the ground is 30∘30^\circ. The angle of elevation changes to 45∘45^\circ after moving 20 meters towards the base of the pole. Then the height of the pole, in meters is

    Correct answer10(3+1)10(\sqrt{3} + 1)

    Why

    At 45∘45^\circ the pole is as far away as it is tall, so call both hh; at 30∘30^\circ the gap is h3h\sqrt{3}.

    The 20 m walk is the difference: h3−h=20h\sqrt{3} - h = 20, so h=203−1h = \frac{20}{\sqrt{3} - 1}.

    Multiply top and bottom by 3+1\sqrt{3} + 1: the pole is 10(3+1)10(\sqrt{3} + 1) m tall.

8 more questions: 1 easy, 4 medium, 3 hard

Sign in free to see all ten and get your score. No payment, no card.

More practice: IPMAT, IIMB, MSAT and NPAT previous year questions

Logos and exam names belong to their owners. Klay Prep is not affiliated with them.