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AFCAT Numerical Ability · Formula sheet

Heights and Distances formulas

14 formulas, 2 reference tables and 32 common traps for AFCAT Numerical Ability Heights and Distances, grouped by subtopic.

Full notes with worked examples

The Trigonometric Ratios at 0°, 30°, 45°, 60° and 90°

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Finding the angle from a value

Reading the table backwards

sin⁡θ=12  ⇒  θ=45∘\sin\theta = \tfrac{1}{\sqrt2} \;\Rightarrow\; \theta = 45^\circ

Complementary angles and the square identity

Complement and square identity

sin⁡(90∘−θ)=cos⁡θ,sin⁡2θ+cos⁡2θ=1\sin(90^\circ - \theta) = \cos\theta, \qquad \sin^2\theta + \cos^2\theta = 1

All the ratios from one given ratio

The third side

hypotenuse2=opposite2+adjacent2\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2

Sine, cosine and tangent at the standard angles

Anglesincostan
0°001100
30°12\tfrac1232\tfrac{\sqrt3}{2}13\tfrac{1}{\sqrt3}
45°12\tfrac{1}{\sqrt2}12\tfrac{1}{\sqrt2}11
60°32\tfrac{\sqrt3}{2}12\tfrac123\sqrt3
90°1100not defined
tan 90° would be 1 divided by 0, which has no value.

Cosecant, secant and cotangent

Anglecosecseccot
0°not defined11not defined
30°2223\tfrac{2}{\sqrt3}3\sqrt3
45°2\sqrt22\sqrt211
60°23\tfrac{2}{\sqrt3}2213\tfrac{1}{\sqrt3}
90°11not defined00

Common traps

√3/2 belongs to 60° for sine

sin⁡60∘=32\sin 60^\circ = \tfrac{\sqrt3}{2} and sin⁡30∘=12\sin 30^\circ = \tfrac12. Swapping them is the most common slip; the memory line puts the bigger value at the bigger angle.

Square the value, not the angle

sin⁡245∘=(12)2=12\sin^2 45^\circ = \left(\tfrac{1}{\sqrt2}\right)^2 = \tfrac12. It is not sin⁡45∘\sin 45^\circ doubled, and it is not the sine of some larger angle.

Cosec goes with sine, sec with cosine

The names mislead: cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta = \tfrac{1}{\sin\theta}, not 1cos⁡θ\tfrac{1}{\cos\theta}. So cosec⁡30∘=2\operatorname{cosec} 30^\circ = 2 and sec⁡30∘=23\sec 30^\circ = \tfrac{2}{\sqrt3}.

Reciprocal is not negative

The reciprocal of 79\tfrac{7}{9} is 97\tfrac{9}{7}. Writing −79-\tfrac{7}{9} or 1−791 - \tfrac{7}{9} confuses flipping with other operations.

Strip the number in front first

2sin⁡θ=12\sin\theta = 1 means sin⁡θ=12\sin\theta = \tfrac12 and θ=30∘\theta = 30^\circ. Reading it as sin⁡θ=2\sin\theta = 2 gives no angle at all, since sine never passes 1.

Flip a reciprocal before looking it up

cot⁡θ=3\cot\theta = \sqrt3 means tan⁡θ=13\tan\theta = \tfrac{1}{\sqrt3}, so θ=30∘\theta = 30^\circ, not 60°. Looking 3\sqrt3 up in the tangent column gives the complement.

Complementary means adding to 90°

sin⁡20∘\sin 20^\circ equals cos⁡70∘\cos 70^\circ, not cos⁡20∘\cos 20^\circ. Check that the two angles add to 90° before swapping sine for cosine.

The identity needs the same angle

sin⁡235∘+cos⁡235∘=1\sin^2 35^\circ + \cos^2 35^\circ = 1, but sin⁡235∘+cos⁡255∘\sin^2 35^\circ + \cos^2 55^\circ is not 1 (it is 2sin⁡235∘2\sin^2 35^\circ).

Put the numbers on the right sides

For cos⁡θ=1517\cos\theta = \tfrac{15}{17} the 15 is the adjacent side and 17 the hypotenuse. Putting 15 opposite the angle swaps sine and cosine in every answer.

The hypotenuse is the longest side

If the side you find comes out longer than the hypotenuse, you added squares where you should have subtracted: the missing shorter side is hyp2−side2\sqrt{\text{hyp}^2 - \text{side}^2}.

One Angle: Towers, Shadows, Ladders and Cliffs

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Height and horizontal distance: the tangent

Height from distance

h=dtan⁡θ,d=hcot⁡θh = d\tan\theta, \qquad d = h\cot\theta

Ladders, ropes, kites and ramps: sine and cosine

A slanting length

h=Lsin⁡θ,x=Lcos⁡θh = L\sin\theta, \qquad x = L\cos\theta

The angle of depression

Depression equals elevation

tan⁡θ=height of observerhorizontal distance\tan\theta = \dfrac{\text{height of observer}}{\text{horizontal distance}}

Finding the angle of elevation

The angle from the sides

θ=tan⁡−1riserun\theta = \tan^{-1}\dfrac{\text{rise}}{\text{run}}

Common traps

Tangent uses the horizontal distance

Tangent links the height to the ground distance. If the length given is the line of sight or a rope, it is the hypotenuse and needs sine or cosine (next concept).

Height over distance, not the reverse

At 60°, a point 40 m away sees a tower 40340\sqrt3 m tall. Writing tan⁡60∘=40h\tan 60^\circ = \tfrac{40}{h} gives 403\tfrac{40}{\sqrt3}, the answer for 30°.

A length is the hypotenuse

The height a 12 m ladder reaches at 60° is 12sin⁡60∘=6312\sin 60^\circ = 6\sqrt3 m. Using 12tan⁡60∘=12312\tan 60^\circ = 12\sqrt3 treats the ladder as lying on the ground.

A broken tree is two pieces

The tree's height is the standing part plus the broken part. Giving the broken part alone, or the standing part alone, is an option set to catch you.

Measure from the horizontal

The angle of depression is between the horizontal and the line of sight. If a question gives the angle with the vertical, subtract it from 90° first.

Move the angle to the ground

Inside the triangle, the angle at the top is the complement of the depression. Using that corner with tangent gives the reciprocal answer. Put the depression at the object instead.

Use the difference of the heights

From the top of an 8 m pole to the top of a 20 m pole, the rise is 12 m. Using 20 m treats the observer as standing on the ground.

Not every angle is standard

If tan⁡θ=27\tan\theta = \tfrac27, no table angle fits, and the right option reads tan⁡−127\tan^{-1}\tfrac27. Do not round it to the nearest table angle.

Two Sightings of One Height: Same Side, Opposite Sides and Speed

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Two points on the same side

Height from the distance walked

h=dcot⁡α−cot⁡βh = \dfrac{d}{\cot\alpha - \cot\beta}

The same set-up with letters

The cotangent difference

cot⁡α−cot⁡β=dh\cot\alpha - \cot\beta = \dfrac{d}{h}

Two points on opposite sides

Opposite sides add

D=h(cot⁡α+cot⁡β)D = h(\cot\alpha + \cot\beta)

Speed from two sightings

Speed

v=h(cot⁡α−cot⁡β)tv = \dfrac{h(\cot\alpha - \cot\beta)}{t}

Common traps

The far point has the smaller angle

Walking towards a tower raises the angle. If the angle falls, the observer moved away, and the distance walked is still the difference of the two distances.

Do not stop at the first distance

Solving for the distance from the near point and giving it as the height is right only at 45°. At any other angle multiply by the tangent.

Distances use cotangent, not tangent

The distance is hcot⁡θh\cot\theta. Subtracting tan⁡α−tan⁡β\tan\alpha - \tan\beta gives a number with no meaning as a length.

Convert to the options' form

1tan⁡α−1tan⁡2α=tan⁡2α−tan⁡αtan⁡αtan⁡2α\tfrac{1}{\tan\alpha} - \tfrac{1}{\tan 2\alpha} = \tfrac{\tan 2\alpha - \tan\alpha}{\tan\alpha\tan 2\alpha}. The right option often prints this form, with the subtraction in the right order on top.

Opposite sides add, same side subtracts

Read for 'opposite sides', 'east and west' or 'on either side' before writing the equation. Subtracting here gives the option for points on the same side.

Both depressions are measured from one top

Two angles of depression from one lighthouse both use the lighthouse's height. Do not use a different height for each ship.

Use the change in horizontal distance

The object moves along the ground or along a level flight path. The lines of sight are slant distances; their difference is not the distance moved.

Check the units of the options

A distance in km covered in tt seconds is a speed of 3600t\tfrac{3600}{t} times that many km/h. Options in m/s and km/h differ by a factor of 3.6.

Two Heights from One Point: Flagstaffs, Two Buildings and Reflections

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A flagstaff on a tower, seen from one point

Two elevations, one distance

H=dtan⁡α,H+h=dtan⁡βH = d\tan\alpha, \qquad H + h = d\tan\beta

Two buildings across a gap

Height seen from a window

H=w+wcot⁡ϕtan⁡θH = w + w\cot\phi\tan\theta

A cloud and its reflection, a balloon and its size

Cloud and reflection

H+hH−h=tan⁡(depression)tan⁡(elevation)\dfrac{H + h}{H - h} = \dfrac{\tan(\text{depression})}{\tan(\text{elevation})}

Common traps

The flagstaff is a difference

The top angle gives the tower plus the flagstaff, not the flagstaff alone. Subtract the tower's height before answering.

One distance for both angles

Both elevations are taken from the same point, so they share dd. Giving each angle its own unknown distance leaves too many unknowns.

Add the observer's height back

The elevation gives only the part of the other building above the observer's eye. The building's height is that part plus the height of the window or roof.

The depression finds the gap

Start with the angle of depression of the foot, which has two known parts (the observer's height and the angle). Starting with the elevation leaves two unknowns.

The reflection is below the water, not the eye

The reflection is HH below the surface, so it is H+hH + h below an eye hh above the water. Using H−hH - h for both makes the two angles equal.

Halve the subtended angle

A balloon subtending 60° puts the centre at rsin⁡30∘\tfrac{r}{\sin 30^\circ}, not rsin⁡60∘\tfrac{r}{\sin 60^\circ}. The line to the centre splits the angle in two.

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