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Chapter 6 · Kerala SSLC Class 10 Maths

Trigonometry

Two ways to learn — a crisp formula reference, or a story that makes angles feel real.

sin / cos / tanStandard valuessin²+cos²=1Heights & distances
Kerala SSLCClass 10MathsChapter 6 — Trigonometry14 min

Real scenarios where trig ratios arise naturally — then the maths.

How story mode works

Each scene is a real situation where trigonometry solves a genuine problem — heights you cannot measure directly, angles you can only observe. The ratio emerges from the situation. Tap any dark panel to see the formal maths.

1

The Lighthouse

A sailor stands on the beach, exactly 100 metres from the base of a lighthouse. He looks up and sees the top of the lighthouse at an angle of elevation of 30°. How tall is the lighthouse?

The sailor cannot climb the lighthouse and drop a tape measure. But he has one angle and one distance. That is enough.

The triangle

The sailor, the base of the lighthouse, and its top form a right triangle. The horizontal distance (100 m) is the adjacent side. The height is the opposite side. The angle at the sailor is 30°.

Which ratio connects opposite and adjacent? That would be tan.

tan 30° = opposite / adjacent

1/√3 = height / 100

height = 100 / √3 = 100√3 / 3 ≈ 57.7 m

The lighthouse is about 57.7 metres tall — measured from the beach with nothing but an angle. That is the power of trigonometry.

2

The Memory Trick

The standard angle values come up in almost every Kerala SSLC trig question. Instead of memorising a table, spot the pattern.

The pattern for sin

Write sin 0°, sin 30°, sin 45°, sin 60°, sin 90° as fractions with denominator 2. The numerators are √0, √1, √2, √3, √4 — in order.

sin 0° = √0/2 = 0

sin 30° = √1/2 = 1/2

sin 45° = √2/2 = 1/√2

sin 60° = √3/2

sin 90° = √4/2 = 1

For cos, read the same pattern backwards: cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0.

tan = sin / cos. You can always derive tan from the sin and cos values rather than memorising a separate tan column.
3

The Shadow Problem

A tree stands 12 metres tall. At a particular time of day it casts a shadow. The shadow is 12√3 metres long. What is the angle of elevation of the sun at that moment?

The tree is vertical — it is the opposite side. The shadow is horizontal — it is the adjacent side. The sun's rays form the hypotenuse. We want the angle θ at the tip of the shadow.

tan θ = opposite / adjacent

= 12 / (12√3)

= 1/√3

We know tan 30° = 1/√3. So θ = 30°.

The insight

We never needed a calculator — the answer came straight from the standard values table. The problem was designed so the numbers matched a known angle.

Exam strategy: Kerala SSLC heights-and-distances problems always use standard angles (30°, 45°, 60°). If your working is giving you an ugly decimal, check whether you used the right ratio — the answer is almost certainly a standard angle.

The Big Picture

Every Kerala SSLC trig question fits one of these patterns. Know these and Chapter 6 is done.

Q

Find a side from angle + another side

Pick sin/cos/tan, substitute, solve

Q

Find an angle from two sides

Compute ratio → match to standard table

Q

Evaluate trig expression at standard angle

Use the √0,√1,√2,√3,√4 pattern

Q

Simplify using identity

sin²θ + cos²θ = 1

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