10. Trigonometry & Complex Numbers

Height and distance

Height and Distance problems form a cornerstone of applied trigonometry, utilizing right-angled triangles, angles of elevation, and depression to determine inaccessible heights and horizontal spans.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Elevation and Depression via Right-Angled Triangles

For any right-angled triangle $\triangle ABC$ with $\angle B = 90^\circ$ and acute angle $\angle BCA = \theta$, the trigonometric ratios define the fundamental relations: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC}$, $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}$, and $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$.

📐 Geometric Interpretation

The angle of elevation is the angle formed by the line of sight with the horizontal upward from the observer, while the angle of depression is formed downward. Geometrically, alternate interior angles equate the angle of depression from an upper point to the angle of elevation from the lower point.

⚙️ Algebraic & Structural Meaning

Translates verbal geometric descriptions into a system of simultaneous algebraic equations involving tangents and cotangents of angles, enabling exact analytical solutions for unknown heights and distances.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Plane Earth Assumption / Vertical Alignment	`Geometric domain (Orthogonal reference)`	Strictly Mandatory / Necessary	Failure to assume perpendicularity between the tower/pole and the horizontal ground invalidates the right-triangle trigonometric framework.
Point-Source Line of Sight	`Idealized optical condition`	Strictly Mandatory	Ignoring observer height when explicitly specified leads to additive height errors in calculations.
Non-zero, Acute Angles of Sight	`\theta \in (0^\circ, 90^\circ)`	Essential for real-valued finite lengths	If $\theta = 0^\circ$ or $90^\circ$, horizontal or vertical collapse occurs, resulting in undefined tangents or zero distances.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Tower and Moving Observer (Angle Doubling)

$$ h = d \cdot \frac{\tan \alpha \tan \beta}{\tan \beta - \tan \alpha} \quad \text{or if } \beta = 2\alpha, \, h = x \tan \alpha \text{ (where } x \text{ is distance moved)}$$

Rule: When an observer moves a distance $d$ towards a tower and the angle of elevation changes from $\alpha$ to $\beta$, use the cotangent difference formula: $h = \frac{d}{\cot \alpha - \cot \beta}.

Application: Frequently asked in UP PGT/TGT where an observer walks towards a building and the angle shifts from $30^\circ$ to $60^\circ$.

Two Points on Opposite Sides of a Tower

$$ h = \frac{d \tan \alpha \tan \beta}{\tan \alpha + \tan \beta} = d \sin \alpha \cos \beta \text{ (when complementary)}$$

Rule: If two points are at distance $d$ apart on a straight line passing through the base of a tower with elevation angles $\alpha$ and $\beta$, the height is given by the product formula.

Cloud and Reflection Problem

$$ h = H \left( \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha} \right) \text{ or } h = x \tan \alpha $$

Rule: If a cloud is at height $h$ above a lake and its reflection in the lake is observed at an angle of depression $\beta$ while the cloud is at elevation $\alpha$, the height of the cloud is calculated directly using the water-surface mirror principle.

Pole on Top of a Building

$$ \frac{H}{h} = \frac{\cot \alpha - \cot \beta_1}{\cot \alpha - \cot \beta_2} $$

Rule: Ratio of heights when a flagpole stands atop a tower, viewed from a common point at varying angles.

Application: High-level problem type in UP PGT screening tests.

Equal Angles from Base to Top

$$ \tan \theta = \frac{h}{d} $$

Rule: Utilize standard Pythagorean triples ($3:4:5$, $5:12:13$) and special angles ($30^\circ, 45^\circ, 60^\circ$) to write down answers without solving quadratic equations.

Rapid Elimination via Tangent Monotonicity

$$ \alpha < \beta \implies \tan \alpha < \tan \beta \implies d_1 > d_2 $$

Rule: Since $\tan \theta$ is strictly increasing on $(0^\circ, 90^\circ)$, closer distance always corresponds to a larger angle of elevation. Instantly eliminate options violating this monotonicity.

Application: Used to discard absurd options in multi-choice questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Ignoring Observer's Eye Height

Common Pitfall: Candidates calculate the height of a tower from the ground level even when the problem states the observer's eye is at a height $h_0$ above the ground.

Counterexample: Observer height = $1.5\text{ m}$, Angle = $45^\circ$, Distance = $10\text{ m}$. Total height $\neq 10\text{ m}$.

Why it fails: The calculated $\tan \theta$ height applies only from the eye level; the total height of the object must include the observer's height ($H = h + h_0$).

🛡️ Exam Defense: Underline statements mentioning 'observer of height $h$' or 'instrument placed at height'.

⚠️ COMMON ERROR #02

Confusing Angle of Elevation with Depression

Common Pitfall: Interchanging angles of elevation and depression or misplacing them in non-horizontal baseline scenarios.

Counterexample: Angle of depression from top of cliff to boat = $30^\circ$; candidate takes angle of elevation from cliff to cloud instead.

Why it fails: Results in inverted triangle proportions and completely incorrect trigonometric ratios.

🛡️ Exam Defense: Always draw a rough horizontal reference line (line of sight parallel to ground) before marking angles.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Single Right Triangle	\theta \in (0, \pi/2)	h = d \tan \theta	Direct scaling of vertical vs horizontal	Single angle and single distance given
Two Points, Same Side	Collinear base points	h = \frac{d}{\cot \alpha - \cot \beta}	Difference of cotangent projections	Observer moves closer/farther by distance d
Two Points, Opposite Sides	Points on opposite sides of base	h = \frac{d}{\cot \alpha + \cot \beta}	Sum of cotangent base spans	Tower stands between two observing stations
Cloud and Reflection	Plane mirror reflection law	h = H \left( \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha} \right)	Symmetric image distance across water surface	Keywords: 'lake', 'cloud', 'reflection', 'depression'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

The angle of elevation of the top of a tower from a point on the ground is $30^\circ$. On walking $50\text{ m}$ towards the tower, the angle of elevation becomes $60^\circ$. Find the height of the tower.

(A) $25\sqrt{3}\text{ m}$(B) $50\sqrt{3}\text{ m}$(C) $25\text{ m}$(D) $\frac{50}{\sqrt{3}}\text{ m}$

📚 Method A: Formal Textbook Derivation

Let height be $h$ and initial distance be $x$. $\tan 30^\circ = \frac{h}{x} \implies x = h\sqrt{3}$. After moving $50\text{ m}$, distance is $x - 50$. $\tan 60^\circ = \frac{h}{x - 50} \implies x - 50 = \frac{h}{\sqrt{3}}$. Substituting $x$: $h\sqrt{3} - \frac{h}{\sqrt{3}} = 50 \implies \frac{2h}{\sqrt{3}} = 50 \implies h = 25\sqrt{3}\text{ m}$.

⚡ Method B: 25-Second Exam Shortcut

Apply shortcut formula: $h = \frac{d}{\cot \alpha - \cot \beta} = \frac{50}{\cot 30^\circ - \cot 60^\circ} = \frac{50}{\sqrt{3} - \frac{1}{\sqrt{3}}} = \frac{50}{\frac{2}{\sqrt{3}}} = 25\sqrt{3}$.

💡 Tip: Memorize standard ratio values for $30^\circ-60^\circ$ triangle shifts: height is always $\frac{\sqrt{3}}{2} \times \text{distance moved}$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The angle of elevation of a cloud from a point $h$ meters above a lake is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$. The height of the cloud above the lake is:

(A) $h \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha}$(B) $h \frac{\tan \beta - \tan \alpha}{\tan \beta + \tan \alpha}$(C) $h \cot \left(\frac{\alpha + \beta}{2}\right)$(D) $h \tan(\alpha + \beta)$

📚 Method A: Formal Textbook Derivation

Let cloud be at height $H$ above water surface. Distance from observation point to cloud level is $H - h$. Depth of reflection below water surface is $H + h$. Using horizontal distance $x = \frac{H-h}{\tan \alpha}$ and $x = \frac{H+h}{\tan \beta}$, equate $x$ to solve for $H$.

⚡ Method B: 25-Second Exam Shortcut

Direct standard textbook result derivation frequently tested as a direct formula in UP PGT.

💡 Tip: Note the plus sign in numerator and minus in denominator for cloud-reflection formulas.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If the angle of elevation of the sun changes from $30^\circ$ to $60^\circ$, the length of the shadow of a pole decreases by $x$ meters. The height of the pole is:

A $\frac{x}{2}$

B $\frac{x\sqrt{3}}{2}$

C $x\sqrt{3}$

D $2x$

Derivation & Analysis:

Let height be $h$. Initial shadow $h\sqrt{3}$, final shadow $\frac{h}{\sqrt{3}}$. Difference is $h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} = x \implies h = \frac{x\sqrt{3}}{2}$.

MCQ 2: A flagstaff $5\text{ m}$ high stands on a building $25\text{ m}$ high. At an observer on the ground, the flagstaff and building subtend equal angles. The distance of the observer from the building is:

A $5\sqrt{30}$

B $10\sqrt{5}$

C $5\sqrt{5}$

D $15\sqrt{2}$

Derivation & Analysis:

Let distance be $x$. Building height $= 25$, flagstaff $= 5$. Total height $= 30$. Using angle sum tangent properties, $\tan(\theta_1 + \theta_2)$ leads to $x = \sqrt{25 \times 30} = \sqrt{750} = 5\sqrt{30}$.

MCQ 3: The upper part of a tree broken by the wind makes an angle of $30^\circ$ with the ground and the distance from the root to the point where the top touches the ground is $10\text{ m}$. The height of the tree is:

A $10\sqrt{3}\text{ m}$

B $\frac{20}{\sqrt{3}}\text{ m}$

C $10(\sqrt{3} + 1)\text{ m}$

D $10\sqrt{3} \text{ m}$

Derivation & Analysis:

Base $b = 10$, angle $30^\circ$. Height of remaining stump $h = 10 \tan 30^\circ = \frac{10}{\sqrt{3}}$. Broken part (hypotenuse) $l = \frac{10}{\cos 30^\circ} = \frac{20}{\sqrt{3}}$. Total height $= \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3}$? Wait: $\frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3}$. Let's verify option B: $\frac{20}{\sqrt{3}} + \frac{10}{\sqrt{3}} = 10\sqrt{3}$. Correct option is (A) $10\sqrt{3}\text{ m}$.

MCQ 4: From the top of a cliff $h$ meters high, the angles of depression of the top and bottom of a tower are $\alpha$ and $\beta$ respectively. The height of the tower is:

A $h\left(1 - \frac{\tan \alpha}{\tan \beta}\right)$

B $h\left(1 - \frac{\tan \beta}{\tan \alpha}\right)$

C $h(1 - \tan \alpha \cot \beta)$

D $h(\cot \alpha - \cot \beta)$

Derivation & Analysis:

Standard derivation yields tower height $H = h\left(1 - \frac{\tan \alpha}{\tan \beta}\right)$.

MCQ 5: A vertical tower is situated on a hill of height $H$. At a point $A$ on the ground, the top and bottom of the tower subtend angles $\alpha$ and $\beta$. The height of the tower is:

A $H(\tan \alpha \cot \beta - 1)$

B $H(\tan \alpha - \tan \beta)$

C $H(\cot \alpha \tan \beta - 1)$

D $\frac{H \sin(\alpha-\beta)}{\cos \alpha \sin \beta}$

Derivation & Analysis:

Standard formula for tower on a hill: $h = H(\tan \alpha \cot \beta - 1)$. Option (A) is correct.

MCQ 6: If the elevation of the sun is $45^\circ$, then the length of the shadow of a $12\text{ m}$ high tree is:

A $6\text{ m}$

B $12\text{ m}$

C $12\sqrt{2}\text{ m}$

D $24\text{ m}$

Derivation & Analysis:

When $\theta = 45^\circ$, $\tan 45^\circ = 1$, so height equals shadow length $= 12\text{ m}$.

MCQ 7: Two ships are sailing in the sea on the two sides of a lighthouse. Their angles of depression from the top of the lighthouse are $30^\circ$ and $45^\circ$. If the height of the lighthouse is $100\text{ m}$, the distance between the ships is:

A $100(1 + \sqrt{3})\text{ m}$

B $100(\sqrt{3} - 1)\text{ m}$

C $200\text{ m}$

D $100\sqrt{3}\text{ m}$

Derivation & Analysis:

Distances are $h \cot 30^\circ = 100\sqrt{3}$ and $h \cot 45^\circ = 100$. Total distance $= 100(1 + \sqrt{3})$. Option (A).

MCQ 8: Assertion (A): The angle of elevation of an object increases as the observer moves closer to the object. Reason (R): The tangent function is strictly increasing in $(0, \pi/2)$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

As distance $d$ decreases, $\frac{h}{d}$ increases, and since $\tan \theta = \frac{h}{d}$, $\theta$ increases because $\tan$ is strictly increasing.

MCQ 9: A pole $10\text{ m}$ high casts a shadow $10\sqrt{3}\text{ m}$ long on the ground. The sun's elevation is:

A $30^\circ$

B $45^\circ$

C $60^\circ$

D $90^\circ$

Derivation & Analysis:

$\tan \theta = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$.

Q10

MCQ 10: If a pole $6\text{ m}$ high casts a shadow $2\sqrt{3}\text{ m}$ long on the ground, then the sun's elevation is:

A $30^\circ$

B $45^\circ$

C $60^\circ$

D $75^\circ$

Derivation & Analysis:

$\tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \implies \theta = 60^\circ$.

Common questions

Always identify the given side (usually the baseline distance or height) and the required unknown. If the relationship involves perpendicular and base, use tangent ($\tan \theta$). If hypotenuse is involved with perpendicular, use sine ($\sin \theta$).

Calculation errors in radical expressions like $\sqrt{3} \approx 1.732$ or failing to account for the observer's height when explicitly stated in the problem statement.

By memorizing standard coefficient ratios for $30^\circ-60^\circ$ and $45^\circ$ triangles (such as $h = \frac{d}{\cot \alpha - \cot \beta}$) and utilizing option elimination via inequality bounds.

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