Is it correct to say: Applying SOH-CAH-TOA to a non-right triangle?

These ratios are defined only inside a right-angled triangle.

Is it correct to say: Reading the wrong side as opposite?

Opposite = the side NOT touching angle θ (and not the hypotenuse). Redraw and label if unsure.

Is it correct to say: Writing sin²θ = sin(θ²)?

sin²θ = (sin θ)². Square the value, not the angle.

Is it correct to say: Mixing sec and cosec?

sec = 1/cos, cosec = 1/sin. Match by first letter opposite: seCant ↔ Cosine, coseCant ↔ Sine.

Is it correct to say: Angle of depression measured from the vertical?

Both elevation and depression are measured from the HORIZONTAL.

Is it correct to say: Ignoring the observer's own height in a heights-and-distances problem?

If the observer's eye-level height is given, add it to the tower height at the end.

Formula sheet (PDF) →

Mathematics · Chapter 6 · Geometry

Trigonometry

Trigonometric ratios, identities, complementary angles, heights and distances — Maharashtra SSC Geometry Ch 6

10 formulas · 3 solved examples · 6 practice questions · Board weight: 6-8 marks · Maharashtra State Board Class 10 · 2025–26 syllabus

Board Exam Tips

→Standard-angle table (0°, 30°, 45°, 60°, 90°) must be memorised — often a 2-mark direct fill-in.
→One 3-mark proof using sin²θ + cos²θ = 1 is a fixed feature of the paper.
→Heights and distances is a 4-mark HOTS — draw diagram with angle of elevation/depression carefully.
→Marathi terms: त्रिकोणमिती (trigonometry), कोन (angle), आधाराचा कोन (angle of elevation).
→Always write 'in right triangle …' before applying SOH-CAH-TOA — earns 1 setup mark.

Formulas(10)

Basic Trigonometric Ratios (SOH-CAH-TOA)Board favourite

sin θ = \frac{Opp}{Hyp}, cos θ = \frac{Adj}{Hyp}, tan θ = \frac{Opp}{Adj}

Symbol	Meaning
$θ$	Angle of interest (0° < θ < 90°)
$O pp$	Side opposite to θ
$A d j$	Side adjacent to θ (not hypotenuse)
$H y p$	Hypotenuse — side opposite the right angle

Reciprocal Ratios

csc θ = \frac{1}{s i n θ}, sec θ = \frac{1}{c o s θ}, cot θ = \frac{1}{t a n θ}

Quotient Identity

tan θ = \frac{s i n θ}{c o s θ}, cot θ = \frac{c o s θ}{s i n θ}

Pythagorean Identity 1Board favourite

sin^{2} θ + cos^{2} θ = 1

Pythagorean Identity 2

1 + tan^{2} θ = sec^{2} θ

Pythagorean Identity 3

1 + cot^{2} θ = csc^{2} θ

Complementary Angle RelationsBoard favourite

sin (9 0^{\circ} - θ) = cos θ, cos (9 0^{\circ} - θ) = sin θ

Standard Angle Values

sin 3 0^{\circ} = \frac{1}{2}, sin 4 5^{\circ} = \frac{1}{2}, sin 6 0^{\circ} = \frac{3}{2}

Angle of Elevation

tan (θ_{e}) = \frac{height}{horizontal distance}

Angle of Depression

tan (θ_{d}) = \frac{height}{horizontal distance}

Solved Examples

1Solved Exampleeasy4 steps

If tan A = 4/3, find all other trigonometric ratios of angle A.

Set opposite = 4k, adjacent = 3k

tan A = \frac{4}{3} \Rightarrow Opp = 4, Adj = 3

2Solved Exampleboard4 steps

Prove that (1 + tan²θ)(1 − sin²θ) = 1.

LHS

(1 + tan^{2} θ) (1 - sin^{2} θ)

3Solved ExampleHOTS4 steps

From a point 50 m away from the foot of a tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower. (Use √3 = 1.732.)

Let height = h, base distance = 50 m, angle = 60°

tan 6 0^{\circ} = \frac{h}{50}

Traps & Common Mistakes

Common Mistakes6

1
Applying SOH-CAH-TOA to a non-right triangle
✓These ratios are defined only inside a right-angled triangle.
2
Reading the wrong side as opposite
✓Opposite = the side NOT touching angle θ (and not the hypotenuse). Redraw and label if unsure.
3
Writing sin²θ = sin(θ²)
✓sin²θ = (sin θ)². Square the value, not the angle.
4
Mixing sec and cosec
✓sec = 1/cos, cosec = 1/sin. Match by first letter opposite: seCant ↔ Cosine, coseCant ↔ Sine.
5
Angle of depression measured from the vertical
✓Both elevation and depression are measured from the HORIZONTAL.
6
Ignoring the observer's own height in a heights-and-distances problem
✓If the observer's eye-level height is given, add it to the tower height at the end.

Practice Yourself

Practice Yourself6 questions

Q1
Find sin 45° + cos 45° in exact form.
Q2
If sin θ = 3/5, find cos θ and tan θ.
Q3
Prove: sec²θ − tan²θ = 1
Q4
A ladder 5 m long leans against a wall making an angle of 60° with the ground. How high up the wall does it reach?
Q5
Find the value of (sin 30° · cos 60°) + (cos 30° · sin 60°).
Q6
Prove: (sin θ + cos θ)² + (sin θ − cos θ)² = 2

Notes

Trigonometry

Trigonometric ratios relate the angles of a right-angled triangle to the ratios of its sides. Because these ratios depend only on the angle (not the triangle size), a right triangle drawn in your notebook can be scaled up to a 100 m tower — same ratios, same trigonometry.

SSC angle

Maharashtra SSC Geometry allots 6-8 marks:

2-mark: direct application of standard-angle values.
3-mark: identity proof (LHS = RHS).
4-mark HOTS: heights and distances — one tower, one observer, one angle.

Standard angle values

Angle	0°	30°	45°	60°	90°
sin	0	1/2	1/√2	√3/2	1
cos	1	√3/2	1/√2	1/2	0
tan	0	1/√3	1	√3	∞

Memory trick: sin increases 0 → 1 across the row, cos decreases 1 → 0 — mirror images.

Identity toolkit

Three Pythagorean identities:

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = csc²θ

Two reciprocal families:

sin ↔ csc, cos ↔ sec, tan ↔ cot
tan = sin/cos, cot = cos/sin

Heights and distances — SSC method

Draw a labelled diagram — mark right angle, angle of elevation/depression, and known length.
Choose the ratio (sin/cos/tan) whose two sides are known/unknown.
Substitute the standard angle value and solve.
Simplify surds if possible; use √3 ≈ 1.732, √2 ≈ 1.414 for numerical answers.

Board vs CBSE

Both boards teach the same identities. Maharashtra SSC likes proofs that combine two or three identities in one line — practise LHS-to-RHS simplifications, and heights-and-distances involving two towers or a moving observer.