What are the side ratios of the 45°-45°-90° and 30°-60°-90° triangles?

A 45°-45°-90° triangle has sides in the ratio 1 : 1 : √2. A 30°-60°-90° triangle has sides in the ratio 1 : √3 : 2 (shortest : medium : longest). These come from cutting a square along its diagonal and cutting an equilateral triangle in half, and they generate all the standard sin/cos/tan values.

How do you find the area of a triangle using sin?

If you know two sides a and b of a triangle and the angle C between them, the area is (1/2) × a × b × sin C. This avoids computing the height separately — the height is derived internally as one side times the sine of the included angle.

What is radian measure and how is it different from degrees?

Radian measure of an angle is its arc length divided by the circle's radius. Degree measure instead divides the arc by the full circumference and multiplies by 360. Both describe the same angle — a full circle is 360° or 2π radians.

How do you find the length of a chord using trigonometry?

The length of a chord equals 2 × r × sin(central angle / 2), where r is the circle's radius. This comes from drawing the perpendicular from the centre to the chord, which bisects both the chord and the central angle.

What is the ratio of sides theorem in a triangle?

In any triangle, the sides are proportional to the sines of their opposite angles: a : b : c = sin A : sin B : sin C. This follows from treating each side as a chord of the triangle's circumcircle, using the fact that a central angle is always twice the inscribed angle on the same arc.

How do you solve heights and distances problems with angle of elevation or depression?

Set up a right triangle using the horizontal distance and the angle of elevation or depression, then use tan(angle) = opposite/adjacent to find the unknown height or distance. Remember to add back the observer's own height, since the angle is measured from eye level, not the ground.

Chapter 6 · Kerala SSLC Class 10 Maths

Trigonometry

Two ways to learn — a crisp formula reference, or a story that builds from two special triangles to chords, side ratios, and real heights-and-distances problems.

Special trianglessin / cos / tanArea = ½ab sinCRadiansChord lengthHeights & distances

Also useful:Chapter 2 — Circles and Angles Practice Problems — Ch 6 (solved questions)1-on-1 Kerala SSLC Maths Tuition

Kerala SSLCClass 10MathematicsChapter 6⏱ 13 min crisp · 20 min story

Six scenes from special triangles to real heights-and-distances problems.

How this works

Six scenes build trigonometry from two special triangles all the way to heights you can't measure directly. Read the story, then tap "The Maths" for the formal result.

Two Special Triangles

Take a square and cut it along a diagonal. Each half is a right triangle with angles 45°, 45°, 90°. Since two angles are equal, the sides opposite them are equal too (isosceles triangles, from Class 8).

If those two equal sides are length 1, how long is the diagonal?

By Pythagoras: diagonal² = 1² + 1² = 2

diagonal = √2

Ratio: 1 : 1 : √2

Any other 45°-45°-90° triangle has these exact same angles, so — by similar triangles — its sides are just this ratio scaled by one number. A short side of 3 cm means a hypotenuse of 3√2 cm.

Now the other special triangle

Take an equilateral triangle with side 2, and cut it in half with a line from one vertex straight down to the midpoint of the opposite side.

That cut is a perpendicular bisector, splitting the equilateral triangle into two right triangles with angles 30°, 60°, 90°. The shortest side is half the original side — so it's 1. The hypotenuse is still the full original side, 2. What's the height?

height² = 2² − 1² = 3

height = √3

So a 30°-60°-90° triangle has sides in the ratio 1 : √3 : 2 — shortest, medium, longest.

Naming the Ratios

Now take ANY angle — not just 30°, 45° or 60°. Mark a point on one side, some distance from the vertex, and drop a perpendicular to the other side.

Slide that point further away or closer in. The height of the perpendicular and the distance to its foot both change — but always in the SAME proportion to the distance from the vertex, because every such triangle has identical angles and is therefore similar (Class 9's Proportion lesson).

height of perpendicular ÷ distance from vertex = a fixed number

distance to foot ÷ distance from vertex = another fixed number

These two fixed numbers are named after the angle itself: the first is the sine of the angle, the second the cosine.

A third ratio compares the perpendicular's height directly to the distance along the OTHER side, not the hypotenuse — this is the tangent.

tan a° = opposite / adjacent

Plug in the two special triangles from Scene 1, and the standard values fall straight out — no memorising required:

sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3

sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1

sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3

For every other angle, mathematicians have measured and tabulated these ratios once and for all — that's the trig table at the end of the textbook (values like sin 40° ≈ 0.6428).

Area Without a Height

Here's a triangle with two known sides, 4 cm and 6 cm, and a 50° angle between them. Normally you would need the height of the triangle to find its area — but you don't have it directly.

Drop a perpendicular from the vertex between the two known sides down to the 6 cm side. That perpendicular is the OPPOSITE side of the 50° angle in a right triangle where the 4 cm side is the hypotenuse.

height = 4 × sin 50°

≈ 4 × 0.7660 = 3.064 cm

Now the area is just half of base times height, using the ordinary formula.

Area = (1/2) × 6 × 3.064

≈ 9.19 cm²

The same idea gives the area of a parallelogram (split it into two triangles) or a rhombus of side s and angle θ (area = s² sin θ) — anywhere you know two sides and the included angle.

A Second Way to Measure Angles

Everything so far has used degrees — but where does "45°" actually come from? Draw a circle centred at the vertex of the angle. The arc trapped between the two sides is some fraction of the whole circle; multiply that fraction by 360, and that is the degree measure.

degree measure = (arc length / circumference) × 360

There is a second, simpler way to measure the same angle: skip the ×360 and the circumference altogether, and just divide the arc length by the radius instead.

radian measure = arc length / radius

Unlike the degree measure, this doesn't depend on the arbitrary number 360 at all — just the circle's own radius and the arc it cuts off.

This is called the radian. The idea was first proposed by the English mathematician Roger Cotes in the eighteenth century; the name "radian" itself came later, from the physicist James Thomson in the nineteenth century.

Chords, Circles, and the Ratio of Sides

New question: given a circle's radius and a chord's central angle, what is the length of the chord itself?

Draw the perpendicular from the centre to the chord — it bisects both the chord and the central angle (an old fact from Circles and Angles).

In the right triangle formed: hypotenuse = radius r

opposite side = half the chord

half the chord = r × sin(central angle / 2)

Use this to find a circumradius. Take an equilateral triangle with sides 3 cm — each side is a chord of its circumcircle, and the central angle of that chord equals twice the angle opposite it in the triangle: 2 × 60° = 120°.

3 = 2r × sin 60°

3 = 2r × (√3/2) = r√3

r = 3/√3 = √3

So the circumradius is √3 cm.

One more idea

If every side of a triangle is a chord, and every chord equals 2r × sin(half its central angle) — and the central angle is always twice the opposite inscribed angle — then every side equals 2r times the SINE of the angle directly opposite it.

side opposite angle A = 2r sin A

side opposite angle B = 2r sin B

side opposite angle C = 2r sin C

One subtlety: if an angle A in the triangle is obtuse, sin(180° − A) = sin A — so the ratio formula works unchanged even then.

Up on the Lighthouse

A person 1.7 m tall stands 10 m from the foot of a tree and sees its top at an angle of elevation of 40°. How tall is the tree?

The person's eye-line is horizontal, 1.7 m off the ground — not at ground level. The angle of elevation is measured from THAT eye-line, so tan 40° only gives the height ABOVE the eyes.

height above eye-line = 10 × tan 40°

≈ 10 × 0.8391 = 8.391 m

total tree height = 8.391 + 1.7 = 10.091 m

About 10.09 metres — and don't forget to add back the observer's own height, a detail that's easy to drop.

A trickier case

A person 1.8 m tall stands atop a 25 m lighthouse and sees a boat at an angle of depression of 35°. How far is the boat from the foot of the lighthouse?

The full vertical height to work with is the lighthouse plus the person: 25 + 1.8 = 26.8 m. The angle of depression, 35°, is measured from the horizontal — so the angle between the VERTICAL line down to the base and the line of sight to the boat is 90° − 35° = 55°.

distance to boat = 26.8 × tan 55°

≈ 26.8 × 1.4281 ≈ 38.27 m

The boat is about 38.27 metres from the foot of the lighthouse.

One more type: sometimes you get TWO angles from TWO positions, and need to solve two equations together.

The canal problem

A boy 1.5 m tall, standing at the edge of a canal, sees the top of a tree on the far bank at an angle of elevation of 70°. Stepping back 10 m, the elevation drops to 25°. How wide is the canal, and how tall is the tree?

Let the canal's width be x, and the tree's height ABOVE the boy's eye-line be y.

From the closer position: y = x tan 70° ≈ 2.7475x

From the farther position: y = (x + 10) tan 25° ≈ 0.4663x + 4.663

Setting them equal: 2.7475x = 0.4663x + 4.663

x ≈ 2.04 m, y ≈ 5.62 m

Width of canal ≈ 2.04 m; tree height = y + 1.5 ≈ 7.12 m, adding back the boy's own height.

The Big Picture

Every idea in this chapter branches from the same two special triangles and the ratios they define. Master these seven moves and Chapter 6 is done.

Special right triangle?

45-45-90 → 1:1:√2, 30-60-90 → 1:√3:2

Define sin/cos/tan for any angle

sin=opp/hyp, cos=adj/hyp, tan=opp/adj

Area from two sides + included angle

Area = (1/2)ab sinC

Measure an angle without degrees

radian = arc length / radius

Length of a chord

chord = 2r sin(central angle / 2)

Compare a triangle's sides

a : b : c = sinA : sinB : sinC

Height you can't measure directly

tan(elevation/depression) = opp/adj, then adjust for eye height

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