Trigonometry in geometry: Examples and Practice

Before this lesson: Trigonometry foundations, Similar triangles, Circles, tangents and power of a point

Your goal: Select a trigonometric relation that simplifies the geometry and check angle domains.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Trigonometry in geometry: the key idea

In a nondegenerate triangle, sides a,b,c are opposite angles A,B,C. The sine rule is a/sin A=b/sin B=c/sin C=2R, where R is the circumradius. The cosine rule is a²=b²+c²−2bc cos A. Area Δ=bc sin A/2=rs=abc/(4R), with inradius r and semiperimeter s. Heron gives Δ²=s(s−a)(s−b)(s−c). Half-angle identities include $\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}$ ; positive roots apply because 0<A/2<90°. Keep one consistent angle unit, and check whether inverse sine produces a second valid angle. Trigonometry is most useful when it removes, rather than multiplies, unknowns.

A 3–4–5 right triangle. The right-angle marker identifies the two legs.

A worked example

Find the inradius and circumradius of a 3–4–5 triangle.

The area is Δ=3·4/2=6 and semiperimeter s=6, so r=Δ/s=1. Also R=abc/(4Δ)=60/24=5/2, as expected from half the hypotenuse of a right triangle.

Your turn: change one thing

In a triangle b=c=5 and A=60°, find a.

Try this on paper before opening the explanation.

Compare your reasoning

Cosine rule gives a²=25+25−50·(1/2)=25, so a=5. The triangle is equilateral.

Methods and connections

Projection, Napier and Mollweide identities

Projection gives a=b cos C+c cos B. From the sine rule and sum-to-product identities, (a−b)/(a+b)=tan((A−B)/2)/tan((A+B)/2), often called the tangent rule or a Napier analogy in this setting. Since (A+B)/2=(180°−C)/2, it can also be written tan((A−B)/2)·tan(C/2). Mollweide’s formulas are (a+b)/c=cos((A−B)/2)/sin(C/2) and (a−b)/c=sin((A−B)/2)/cos(C/2). Derive them by replacing a,b,c with 2R sin A,2R sin B,2R sin C, then applying sum-to-product formulas; the half-angle denominators are positive.

Heron and the half angles

From cos A=(b²+c²−a²)/(2bc), the identity sin²(A/2)=(1−cos A)/2 yields (a²−(b−c)²)/(4bc)=(s−b)(s−c)/(bc). Similarly cos²(A/2)=s(s−a)/(bc). Multiply these and use sin A=2sin(A/2)cos(A/2) with Δ=bc sin A/2 to obtain Heron’s formula. This connects the formulas instead of treating each as an isolated rule.

Exradii and centres

The excircle opposite A touches BC and the extensions of AB,AC. Its radius is rₐ=Δ/(s−a). The area identity follows by subtracting the triangle on BC from the two triangles on the extended adjacent sides, giving Δ=rₐ(b+c−a)/2. In the 3–4–5 triangle, the exradius opposite side 5 is 6/(6−5)=6. The circumcentre measures vertex distance, whereas incentre and excentres measure distance to side lines.

Quadrilateral area from diagonals

For a convex quadrilateral with diagonals of lengths p,q meeting at angle θ, area is pq sin θ/2. Split into the four triangles at the diagonal intersection and add their half-product-sine areas. Opposite intersection angles have the same sine. For perpendicular diagonals, area reduces to pq/2; equal diagonal lengths are not required.

Regular polygons

A regular n-gon with circumradius R splits into n isosceles triangles with central angle 360°/n. Its side length is 2R sin(180°/n), apothem R cos(180°/n), and area nR² sin(360°/n)/2. For a regular hexagon, the side is R and area is $\frac{3\sqrt3 R^2}{2}$ . Use radians consistently if your calculator or formula uses 2π/n instead.

Pause and check

A trap to avoid: Ignoring sine ambiguity or applying a triangle-centre formula with inconsistent notation.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove Δ=rs by splitting the triangle at the incentre.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Join the incentre I to all three vertices. Its perpendicular distance to each side is the common inradius r. The three triangle areas sum to Δ=(ar+br+cr)/2=r(a+b+c)/2=rs. The incentre lies inside the triangle, so these three regions partition it.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. In the sine rule, a/sin A equals what?

Foundation

Hint

R denotes circumradius.

Answer and reasoning

2R. The extended sine rule is a=2R sin A.

2. In Δ=rs, s denotes what?

Foundation

Side a
Circumradius
Square root of area
Semiperimeter

Hint

Use s=(a+b+c)/2.

Answer and reasoning

Semiperimeter. It is half the perimeter.

3. What is the inradius of a 3–4–5 triangle?

Core

Hint

Use area 6 and semiperimeter 6.

Answer and reasoning

1. r=Δ/s=1.

4. What is the circumradius of a 3–4–5 triangle?

Core

Hint

A right triangle’s hypotenuse is a diameter.

Answer and reasoning

5/2. R=5/2, also abc/(4Δ)=60/24.

5. If b=c=5 and A=60°, what is a?

Stretch

10
$\sqrt{75}$
5
$5\sqrt{2}$

Hint

Use the cosine rule.

Answer and reasoning

5. a²=25+25−2·25·1/2=25.

6. Why can inverse sine create two triangle candidates?

Stretch

sin θ=sin(180°−θ)
Sine is always negative
Angles do not sum to 180°
Every triangle is right

Hint

Check both the acute and supplementary possibilities.

Answer and reasoning

sin θ=sin(180°−θ). The remaining angle sum and side constraints determine which candidates are valid.

7. Heron’s formula uses sides and which auxiliary quantity?

Foundation

Only a right angle
The number of vertices
The semiperimeter
The longest diagonal

Hint

Let s=(a+b+c)/2.

Answer and reasoning

The semiperimeter. The area is $\sqrt{s(s-a)(s-b)(s-c)}$ .

8. A triangle has area 24 and semiperimeter 12. Find its inradius.

Core

Hint

Use Δ=rs.

Answer and reasoning

2. r=24/12=2.

9. Why take positive roots in the half-angle formulas for a triangle?

Stretch

Its half-angles lie strictly between 0° and 90°
Every angle is 90°
Square roots are always negative
The triangle may have zero angles

Hint

Sine and cosine are positive in that interval.

Answer and reasoning

Its half-angles lie strictly between 0° and 90°. A nondegenerate triangle has 0<A<180°, so both sin(A/2) and cos(A/2) are positive.

Choose your next step

Continue to Constructing triangles. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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