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GCSE Maths · Geometry and measures

Area of a triangle using sine GCSE Questions and Worked Answers

Use A = ½ab sin C when two sides and the angle between them are known. The sine supplies the perpendicular height that the ordinary triangle-area formula needs.

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Start with the meaning

What you need to know about area of a triangle using sine

A triangle's area measures the amount of flat surface inside it. If you know its base and perpendicular height, the area is half their product: two matching triangles fill a parallelogram. Sometimes a question gives a sloping side instead of the height. Sine lets us recover the height from that side and the angle it makes with the base.

See the idea first

Replace the missing height with an equal expression

Call the base a, the other given side b, and the angle between them C. Drop a perpendicular height h. In the small right triangle, sin C = h/b, so h = b sin C. The area is ½ × a × h; replacing h gives A = ½ab sin C. A names the area, while C names the included angle.

a (base)bhC
Drop a perpendicular height to the base. For this acute angle C, sin C = h/b, so h = b sin C. Substituting that height into half base times height gives the area formula.
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Find area

What the problem asks: Two sides of 7 cm and 10 cm enclose 30°. Find the area.

How to solve it: A = ½ × 7 × 10 × sin 30° = 17.5 cm². Use square units because the result measures surface.

Find a side

What the problem asks: Area is 18 cm², one side is 9 cm and its included angle with the missing side is 30°. Find that side.

How to solve it: 18 = ½ × 9 × b × 0.5, so 18 = 2.25b and b = 8 cm.

Find an angle

What the problem asks: Area is 12 cm² and the two sides enclosing C are 6 cm and 8 cm. Find possible C.

How to solve it: 12 = ½ × 6 × 8 × sin C = 24 sin C, so sin C = 12 ÷ 24 = 0.5. Both 30° and 150° are possible unless extra information rules one out.

A reliable routine

Use two sides and their included angle

The sine-area formula applies to any non-degenerate triangle when the selected angle lies between the selected sides. For an obtuse included angle, an external perpendicular gives the same height because sin C = sin(180° − C).

  1. Mark the two sides and the angle where they meet.
  2. Use degree mode; write A = ½ab sin C before substituting.
  3. For a missing side or angle, rearrange the area equation.
  4. Keep unrounded calculator values until the end; check square units and any acute/obtuse restriction.

Check: If you know three sides but no angle, first use the cosine rule. If C = 90°, sin C = 1 and the formula becomes the ordinary right-triangle area formula.

Fully worked

Area of a triangle using sine GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Direct area

Higher only3 marks
Question

Sides 8 cm and 11 cm enclose 30°. Find the area.

A=12(8)(11)sin⁡30∘A=\tfrac12(8)(11)\sin30^\circ =44(0.5)=44(0.5) =22 cm2=22\text{ cm}^2

The angle belongs between the two sides.

Example 2

Calculator area

Higher only3 marks
Question

Sides 9 m and 12 m enclose 50°. Find area to 3 significant figures.

A=12(9)(12)sin⁡50∘A=\tfrac12(9)(12)\sin50^\circ =41.366399…=41.366399\ldots A=41.4 m2A=41.4\text{ m}^2

Use degree mode.

Example 3

Obtuse angle

Higher only3 marks
Question

Sides 5 cm and 16 cm enclose 150°. Find area.

The included angle may be obtuse.

A=12(5)(16)sin⁡150∘A=\tfrac12(5)(16)\sin150^\circ =40(0.5)=20 cm2=40(0.5)=20\text{ cm}^2
Example 4

Missing side

Higher only4 marks
Question

A triangle has area 27 cm². A side of 12 cm meets the missing side b at 30°. Find b.

27=12(12)bsin⁡30∘27=\tfrac12(12)b\sin30^\circ 27=3b27=3b b=9 cmb=9\text{ cm}

Divide by the entire coefficient of b.

Example 5

Possible angles

Higher only4 marks
Question

Area is 15 cm². Sides 6 cm and 10 cm enclose C. Find both possible angles.

15=12(6)(10)sin⁡C15=\tfrac12(6)(10)\sin C sin⁡C=15/30=0.5\sin C=15/30=0.5

Therefore C = 30° or 150°. Either gives the same perpendicular height.

Example 6

Combined area

Higher only4 marks
Question

A quadrilateral is split by a diagonal into two triangles with no overlap. One has sides 6 cm and 8 cm enclosing 30°; the other has base 8 cm and perpendicular height 5 cm. Find total area.

A1=12(6)(8)sin⁡30∘=12A_1=\tfrac12(6)(8)\sin30^\circ=12 A2=12(8)(5)=20A_2=\tfrac12(8)(5)=20 A=12+20=32 cm2A=12+20=32\text{ cm}^2

The diagonal separates the two interiors.

10 original questions · total 30 marks

Area of a triangle using sine GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 35 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
1

Area

Higher only3 marks

Find area for sides 4 cm and 9 cm enclosing 30°.

Show worked answer
A=12(4)(9)(0.5)=9 cm2A=\tfrac12(4)(9)(0.5)=9\text{ cm}^2
2

Right angle

Higher only2 marks

Find area for sides 7 m and 10 m enclosing 90°.

Show worked answer
A=12(7)(10)(1)=35 m2A=\tfrac12(7)(10)(1)=35\text{ m}^2
3

Exact area

Higher only3 marks

Find exact area for sides 4 cm and 6 cm enclosing 60°.

Show worked answer
A=12sin⁡60∘A=12\sin60^\circ =12×32=63 cm2=12\times\frac{\sqrt3}{2}=6\sqrt3\text{ cm}^2
4

Obtuse area

Higher only3 marks

Find area for sides 8 cm and 13 cm enclosing 150°.

Show worked answer
A=12(8)(13)(0.5)=26 cm2A=\tfrac12(8)(13)(0.5)=26\text{ cm}^2
5

Side

Higher only4 marks

Area is 20 cm². Sides a and 8 cm enclose 30°. Find a.

Show worked answer
20=12a(8)(0.5)20=\tfrac12 a(8)(0.5) 20=2a20=2a a=10 cma=10\text{ cm}
6

Acute angle

Higher only4 marks

Area is 10 cm². Sides 5 cm and 8 cm enclose an acute angle C. Find C.

Show worked answer
10=20sin⁡C10=20\sin C sin⁡C=0.5\sin C=0.5

The acute solution is 30°; 150° is excluded.

7

Maximum area

Higher only3 marks

Sides 6 cm and 9 cm enclose a variable angle. What is the largest possible area?

Show worked answer

The largest sine value is 1, at 90°.

Amax⁡=12(6)(9)=27 cm2A_{\max}=\tfrac12(6)(9)=27\text{ cm}^2
8

Impossible data

Higher only3 marks

Can sides 5 cm and 8 cm enclose a triangle of area 25 cm²?

Show worked answer

The maximum area is ½ × 5 × 8 = 20 cm². Alternatively sin C would be 25/20 = 1.25, impossible. No.

9

Choose the angle

Higher only2 marks

In triangle ABC, AB = 7 cm, AC = 9 cm and angle BAC = 40°. Write the area calculation.

Show worked answer

AB and AC meet at A, so use angle BAC.

Area=12×7×9×sin⁡40∘\text{Area}=\tfrac12\times7\times9\times\sin40^\circ
10

Scale the sides

Higher only3 marks

Both sides enclosing an unchanged angle double. How does area change?

Show worked answer
12(2a)(2b)sin⁡C=4(12absin⁡C)\tfrac12(2a)(2b)\sin C=4\left(\tfrac12ab\sin C\right)

The area becomes four times as large.

Examiner-style feedback

Common area of a triangle using sine mistakes

Using an opposite angle

The angle must be between the two sides in the product.

Missing the second angle

Inverse sine alone may miss an obtuse solution.

Using a sloping height in ½bh

A height must be perpendicular; this is why sine is needed.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. Mark the included angle.
  2. Recover height using sine.
  3. Keep exact values until rounding.
  4. Check angle restrictions.
Quick answers

Area of a triangle using sine FAQ

Does it work for obtuse triangles?

Yes. Use the included angle; its sine is positive between 0° and 180°.

Is this Higher content?

Yes, the sine area formula is G23 Higher in AQA GCSE Maths.

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What to revise next

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Content standards

Curriculum and rights review

G23 Higher: trigonometric triangle area, reverse problems and ambiguous angles. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references