Some Applications of Trigonometry — Detailed Notes

Chapter 9 | CBSE Class 10 Mathematics (NCERT)

Study Diagrams (Visual Revision)

Use these board-style figures while reading the chapter. Open the Diagrams tab for a larger gallery.

Heights & Distances — Decision Map

Elevation & Depression Figure

Heights & Distances — Formula Patterns

9.1 Heights and Distances

Key definitions

Line of sight: Line from the observer’s eye to the point viewed on the object.
Angle of elevation: Angle between line of sight and horizontal when the object is above horizontal (looking up).
Angle of depression: Angle between line of sight and horizontal when the object is below horizontal (looking down).

Basic method

Draw a clear figure; mark right angles and given lengths/angles.
Identify the right triangle(s) and the unknown.
Choose $sin$ , $cos$ , or $tan$ linking known and unknown sides.
Solve; interpret the answer (add observer height if needed).

For a tower $A B$ viewed from $C$ on level ground with elevation $∠ A C B = θ$ : $tan θ = \frac{A B}{B C} .$

Worked ideas (NCERT examples)

Example 1 — Simple tower: Distance $15$ m, elevation $6 0^{\circ}$ . $tan 6 0^{\circ} = \frac{h}{15} \Rightarrow h = 153 m .$

Example 2 — Ladder on pole: Pole $5$ m; repair point $1.3$ m below top $\Rightarrow$ climb height $3.7$ m; ladder at $6 0^{\circ}$ to ground. $sin 6 0^{\circ} = \frac{3.7}{ladder} \Rightarrow ladder \approx 4.28 m, cot 6 0^{\circ} = \frac{base}{3.7} \Rightarrow base \approx 2.14 m .$

Example 3 — Observer height: Observer $1.5$ m tall, $28.5$ m from chimney, elevation of top $4 5^{\circ}$ from eyes. $tan 4 5^{\circ} = \frac{A E}{28.5} = 1 \Rightarrow A E = 28.5 \Rightarrow chimney = 30 m .$

Example 4 — Building + flagstaff: Building $10$ m; elevations of top of building and flagstaff $3 0^{\circ}$ and $4 5^{\circ}$ . $A P = 103, 10 + x = 103 \Rightarrow x = 10 (3 - 1) \approx 7.32 m .$

Example 5 — Sun altitudes $6 0^{\circ}$ and $3 0^{\circ}$ : Shadow lengthens by $40$ m. $h = x 3, \frac{h}{x + 40} = \frac{1}{3} \Rightarrow x = 20, h = 203 m .$

Example 6 — Depressions of top and bottom: From a multi-storeyed building, angles of depression of the top and bottom of an $8$ m building are $3 0^{\circ}$ and $4 5^{\circ}$ . Using alternate angles with the horizontal through the top: $P D + 8 = P D 3 \Rightarrow P D = \frac{8}{3 - 1} = 4 (3 + 1) .$ Height of multi-storeyed building $= P D + 8 = 43 + 12$ m; distance between buildings $= 4 (3 + 1)$ m.

Example 7 — River width: Bridge height $3$ m; depressions of opposite banks $3 0^{\circ}$ and $4 5^{\circ}$ . $A D = 33, B D = 3, A B = 3 (1 + 3) m .$

Better Method / Strategy:

Default to $tan$ for height–distance on level ground.

Use $sin$ for rope/string/ladder (hypotenuse).

Two angles $\Rightarrow$ introduce one auxiliary ( $x$ or $h$ ), write two equations, eliminate.

Depression: draw horizontal through eye; mark alternate interior angle equal to depression.

Typical board structures

Structure	Approach
Single elevation	One $tan$ equation
Observer not at ground level	Split height = above eye + eye height
Two elevations from same point	Two triangles sharing base; subtract heights
Move towards object ( $3 0^{\circ} \to 6 0^{\circ}$ )	Two positions; difference of distances = walk length
Shadow lengthens	Two sun altitudes; difference of shadow lengths given
Depressions of two points	Alternate angles; often $4 5^{\circ}$ makes base = height

9.2 Summary (NCERT)

Line of sight; angle of elevation; angle of depression — as defined above.
Heights/distances of objects can be found using trigonometric ratios.

Figure Checklist (Before Algebra)

Vertical object drawn straight; ground horizontal.
Observer eye marked if height given.
Line of sight drawn; angle labelled at the correct vertex.
Right angle marked at the foot of the vertical.
For depression: extra horizontal through the eye, with alternate angle shown.
Unknowns named ( $h$ , $x$ , $ℓ$ ) on the figure.

Algebra Patterns Worth Memorising

A. Single tower

$h = d tan θ$

B. Walk from $3 0^{\circ}$ to $6 0^{\circ}$ (height $h$ above eye)

Distances $h 3$ and $\frac{h}{3}$ ; walk $= \frac{2 h}{3}$ .

C. Building $H$ + tower $h$ , elevations $4 5^{\circ}$ and $6 0^{\circ}$ from same point

Base $= H$ (from $4 5^{\circ}$ ); then $H + h = H 3$ $\Rightarrow$ $h = H (3 - 1)$ .

D. Sun altitudes $6 0^{\circ}$ and $3 0^{\circ}$ , shadow grows by $m$

$h = 203 when m = 40 (scale: h = \frac{m}{2} 3 in that special case) .$ General: $h = x 3$ , $\frac{h}{x + m} = \frac{1}{3}$ $\Rightarrow$ $x = \frac{m}{2}$ , $h = \frac{m 3}{2}$ .

E. Two depressions $3 0^{\circ}$ and $4 5^{\circ}$ from height $H$ to opposite banks

Width $= H (1 + 3)$ .

F. Lighthouse, two ships same side, depressions $3 0^{\circ}$ and $4 5^{\circ}$ , height $H$

Distances $H 3$ and $H$ ; separation $= H (3 - 1)$ .

Word-Problem Translation Table

Phrase	Geometry
“Angle of elevation of the top”	Angle at ground/eye up to top
“Inclined at $6 0^{\circ}$ to the ground”	Angle between ladder/rope and ground
“Angle of depression of the foot”	From top, looking down to foot
“At the same time” (shadows)	Same sun altitude for both objects
“No slack in the string”	String = straight hypotenuse

Exam-Oriented Analysis

Aspect	Detail
Weight	Usually one solid 3–4 mark application
Question types	Tower; kite/string; slides; walking towards building; statue on pedestal; ships from lighthouse; canal
Common mistakes	Forgetting observer height; mixing elevation/depression; using $sin$ where $tan$ needed; degree vs ratio errors; not drawing parallel horizontal for depression
Scoring tips	Neat figure with right angle marked; write ratio equation with reason; keep exact surds unless approximation asked

Scoring tip: Label what you find ( $h$ , $x$ ) on the figure; examiners award method marks even if arithmetic slips.

Method chooser

Draw figure → mark θ
Hypotenuse unknown/known (rope/ladder)? → sin/cos
Otherwise height–distance? → tan
Two angles? → two equations → eliminate
Depression? → alternate angle → then tan

Fully Worked Skeleton (Statue on Pedestal Style)

Setup: Statue height $s$ on pedestal height $p$ . From a ground point, elevation of statue top is $6 0^{\circ}$ , of pedestal top is $4 5^{\circ}$ .

Figure: Pedestal $B C = p$ , statue $A B = s$ , observer at $D$ ; $∠ B D C = 4 5^{\circ}$ , $∠ A D C = 6 0^{\circ}$ , base $D C = x$ .

Equations: $tan 4 5^{\circ} = \frac{p}{x} = 1 \Rightarrow x = p$ $tan 6 0^{\circ} = \frac{p + s}{x} = 3 \Rightarrow p + s = p 3 \Rightarrow s = p (3 - 1)$ If $s = 1.6$ , then $p = \frac{1.6}{3 - 1} = 0.8 (3 + 1)$ after rationalising.

This skeleton matches many “object on top of object” board questions.

Timing Plan (4-mark question)

Minute	Action
0–1	Read twice; sketch figure; mark data
1–3	Write trig equations; solve
3–4	Rational units; quick substitution check

Final check: Does the larger angle correspond to the larger opposite side? If elevation rose, did the observer move closer?