Quick Revision

Elevation and Depression

The line of sight joins the eye to the object. Both angles are measured from the horizontal through the eye.
Object above the eye: angle of elevation. Object below the eye: angle of depression.
The depression of B seen from A equals the elevation of A seen from B (alternate angles), so mark it at the lower point.
The right angle is at the foot of the tower.

$Equal angles of elevation and depression on one line of sight$

$30^\circ$ - $60^\circ$ - $90^\circ$ : sides $1 : \sqrt{3} : 2$ , with the shortest side opposite $30^\circ$ .
$45^\circ$ - $45^\circ$ - $90^\circ$ : sides $1 : 1 : \sqrt{2}$ .
$h = d\tan\theta$ : at $30^\circ$ , $h = \frac{d}{\sqrt{3}}$ ; at $45^\circ$ , $h = d$ ; at $60^\circ$ , $h = \sqrt{3}d$ .
Line of sight: $2h$ at $30^\circ$ , $\sqrt{2}h$ at $45^\circ$ , $\frac{2h}{\sqrt{3}}$ at $60^\circ$ .
Height equal to distance means $45^\circ$ ; distance $\sqrt{3}$ times the height means $30^\circ$ .

For example: 15 m from the foot at $60^\circ$ , the tower is $15\sqrt{3}$ m tall.

Eye height e: use $h - e$ in the triangle, then add e back for the full height.
A ratio given: $\tan\theta = \frac{3}{4}$ makes height : distance : line of sight $= 3 : 4 : 5$ .
Other angles come with their values in the question: $\tan 15^\circ = 2 - \sqrt{3}$ , $\tan 75^\circ = 2 + \sqrt{3}$ .
For decimals use $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$ , at the last step only.
An angle given from the vertical: the elevation or depression is $90^\circ$ minus it.

Shadow $= h\cot\theta$ , where $\theta$ is the sun's altitude.
At one moment, height : shadow is the same for every object.
$60^\circ$ to $30^\circ$ : the shadow triples, growing by $\frac{2h}{\sqrt{3}}$ . From $60^\circ$ to $45^\circ$ , and again from $45^\circ$ to $30^\circ$ , it becomes $\sqrt{3}$ times as long.
$45^\circ$ to $30^\circ$ : it grows by $h(\sqrt{3} - 1)$ .
[Tier 2] Lamp at height L, person of height p at distance x from the post, shadow s: $\frac{p}{s} = \frac{L}{x + s}$ .

[SSC Tip] The sun's altitude is an angle of elevation, measured at the tip of the shadow.

Ladder L at $\alpha$ to the ground: foot $L\cos\alpha$ from the wall, top $L\sin\alpha$ up. Across a street: width $L(\cos\alpha + \cos\beta)$ .
A sliding ladder keeps its length (Pythagoras).
A kite string or a wire is the hypotenuse; measure the height from the hand.
Wire between two poles: the difference of their heights is the upright side.
Broken tree: standing part + broken part = tree. At $30^\circ$ with the ground, the tree was $\sqrt{3}d$ .
Road or ramp at $\theta$ : rise = length $\times \sin\theta$ .

$Two points on one side of a tower at angles alpha and beta$

Same side, d apart: $d = h(\cot\alpha - \cot\beta)$ . Opposite sides: $d = h(\cot\alpha + \cot\beta)$ .
$30^\circ$ to $60^\circ$ : the walk is twice what is left, and the height is $\frac{\sqrt{3}}{2}$ of the walk.
Complementary angles from distances a and b: $h = \sqrt{ab}$ .
Boats seen from a cliff: the same figure, with each depression moved down to its boat.

Speed = distance between the two positions divided by the time.
Aeroplane at height h, $60^\circ$ to $30^\circ$ on one side: it flies $\frac{2h}{\sqrt{3}}$ . Passing overhead: add the two cots.
Balloon rising from a point d away: it rises $d(\tan\beta - \tan\alpha)$ .
Time still needed to reach the foot: $\frac{t}{2}$ for $30^\circ$ to $60^\circ$ , and $\frac{\sqrt{3} + 1}{2}t$ for $30^\circ$ to $45^\circ$ or $45^\circ$ to $60^\circ$ .
km/h $\times \frac{5}{18}$ gives m/s; m/s $\times \frac{18}{5}$ gives km/h.

$Tower split by the eye line of a building into two parts$

From a building h tall: $\frac{H - h}{h} = \frac{\tan\alpha}{\tan\beta}$ . Elevation $60^\circ$ with depression $30^\circ$ gives $H = 4h$ .
Statue on a pedestal, one point: pedestal $\div$ whole $= \tan\alpha \div \tan\beta$ . For $30^\circ$ and $60^\circ$ the pedestal is half the statue.
Two towers facing each other: difference of heights $= d\tan\theta$ .
A point between two poles: its two distances add up to the gap.

[Tier 2] Cloud H above a lake, seen from h above the water, at elevation $\alpha$ , with its reflection at depression $\beta$ : $H - h = x\tan\alpha$ and $H + h = x\tan\beta$ , so $H = h\frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}$ . For $30^\circ$ and $60^\circ$ , $H = 2h$ .
[Tier 2] Height from two points d apart on one side: $h = \frac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha}$ .
Check any answer in $\alpha$ and $\beta$ by putting $\alpha = 30^\circ$ and $\beta = 60^\circ$ , and then a second pair such as $30^\circ$ and $45^\circ$ , because two different formulas can agree at one pair.

Draw the figure; mark the right angle, the given angle and the known side.
Slips to watch: tan and cot swapped ( $h\sqrt{3}$ for $\frac{h}{\sqrt{3}}$ ), the eye height left out, the gap between two points used as a distance from the foot, a depression marked from the vertical, the line of sight given as the distance, and km/h mixed with m/s.
Check: below $45^\circ$ the height is less than the distance.

About one CGL 2025 Tier 1 shift in two or three had a question from this chapter. Finish with the PYQ Practice Sets.