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This page sets out the method for area of any triangle in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Area of any triangle topic page.
When you see a triangle with two side lengths and the angle where those sides meet, and no right angle.
When the area and two sides are given and you need the angle between them.
Area = 1/2 × side × side × sin(included angle).
Sin 30 = 1/2, so 1/2 × 8 × 10 × 1/2 = 20.
Sin 60 = √3/2, so equal sides of 6 give area 9√3.
Sin 45 = √2/2, so the area keeps a factor of √2.
Sin 90 = 1, so the formula becomes the ordinary 1/2 × base × height when the angle is right.
Leave √3 and √2 in the answer when the angle is 60 or 45.
Sketch the triangle and mark the two sides you know. The angle you need is the one in the corner where those sides meet. Call that angle C, and call the sides a and b. The letters do not have to match a printed diagram. What matters is that C is between a and b. If the diagram gives you angle A between sides b and c, the area is 1/2bcsin A. Same formula, different letters.
A side and the angle opposite it is a different method. You will see that layout when one angle and the side facing it are the labelled pair, and the second side is not enclosing that angle. Stop and do not force 1/2absin C onto it. Also check the units. Sides in centimetres give an area in cm^2. If one side is in metres and the other in centimetres, convert first so both lengths use the same unit.
Method: 1. Mark the angle that sits between the two known sides. 2. Use 1/2absin C with that included angle. 3. If the letters differ, write 1/2bcsin A for the angle between sides b and c.
Multiply in one calculator line: 0.5 × a × b × sin C. For sides 8 cm and 10 cm with included angle 30^°, sine of 30^° is exactly 1/2, so the area is 1/2 × 8 × 10 × 1/2 = 20 cm^2. You do not need a decimal for that angle. For sides 7 cm and 9 cm with included angle 48^°, the calculator value of 1/2 × 7 × 9 × sin 48^° is 23.4 cm^2 to 1 decimal place.
Keep the calculator in degrees. Sine of 48 in radians is a different number and the area will not match a sketch. Do not round sin 48^° to 0.7 and then multiply. Type the whole product and round the area at the end. If the question asks for 3 significant figures, 23.409ldots would be 23.4, which already has three significant figures. Read the demand. One decimal place and three significant figures are not always the same instruction.
Method: 1. Multiply 0.5× a× b×sin C in one calculator line, in degrees. 2. Sides 8cm and 10cm with 30^° give 1/2× 8× 10×1/2=20cm^2. 3. Sides 7cm and 9cm with 48^° give 23.4cm^2 to 1 decimal place.
If the area is known and the two sides are known, rearrange to sin C = frac2 × areaab. For an area of 30 cm^2 between sides 8 cm and 10 cm, sin C = frac301/2 × 8 × 10 = 30/40 = 0.75. Inverse sine of 0.75 is 48.6^° to 1 decimal place. That is the acute answer.
Sine is also positive for an obtuse angle. The other possibility is 180^° - 48.6^° = 131.4^°. Both angles sit between the sides of 8 cm and 10 cm, and both give area 30 cm^2, because sin 131.4^° = sin 48.6^°. Unless the question says the angle is acute, or the diagram is clearly acute, give both. An included angle of 0^° or 180^° would flatten the triangle, so any angle strictly between them is possible here.
Method: 1. Rearrange to sin C=dfrac2×areaab. 2. Area 30cm^2 between sides 8cm and 10cm gives sin C=0.75. 3. The acute angle is 48.6^°, and the obtuse possibility is 180-48.6=131.4^°.
Once you know the area, you can find a perpendicular height. If the area is 1/2 × 13 × h, then h = frac2 × area13. With sides 13 cm and 11 cm and included angle 67^°, the area is 1/2 × 13 × 11 × sin 67^° = 65.8 cm^2 to 3 significant figures. The height to the side of 13 cm is 11sin 67^°, because the 13 cancels, and that is 10.1 cm to 1 decimal place.
Use the full calculator value of the area if you divide to find the height, or use the cancelled form h = 11sin 67^° so you never round early. Rounding the area to 65.8 and then dividing still gives 10.1 cm here, but that agreement is lucky. The safe habit is one calculator line for the area, and a fresh line for the height from the original sine. Write the units: cm^2 for area, cm for the height.
Method: 1. Find the area first, or cancel straight to the height. 2. Sides 13cm and 11cm with 67^° give area 65.8cm^2 to 3 significant figures. 3. The height to the 13cm side is 11sin 67^°=10.1cm to 1 decimal place.
The formula tfrac12absin C only works when C sits between the two sides a and b. In an exam diagram the angle is often drawn at one corner and two lengths are written on the sides that meet there. If the angle you are given is at a different corner, the question has hidden a step. Find the missing angle first using angles in a triangle add to 180^°.
A reliable habit is to put your finger on the angle and slide along the two sides that touch it. Those two sides are your a and b. If the lengths printed are on other sides, you are not ready to use the formula yet. Real questions usually say "work out the area of the triangle" and leave the structure to you, so the first line of working should name the two sides and the included angle.
Method: 1. Mark the angle and the two sides that meet at it. 2. If the given angle is not between the given sides, find the missing angle from 180^°. 3. Write Area = tfrac12 × a × b × sin C with the numbers in. 4. Calculate, round at the end, and state the unit as cm^2 or m^2.
Tip: The two sides and the angle must be touching. If they are not, find another angle before you start.
On a non-calculator paper the angle is nearly always 30^°, 45^°, 60^°, 120^° or 150^°. Learn the values sin 30^° = tfrac12, sin 45^° = tfrac√(2)2 and sin 60^° = tfrac√(3)2. An obtuse angle uses the same value as its partner: sin 150^° = sin 30^° and sin 120^° = sin 60^°, because the sine graph is symmetrical about 90^°.
The question will say "give your answer in the form a√(3)" or "give your exact answer". Keep the surd and do not turn it into a decimal. For sides 8 and 6 with an angle of 60^° the area is tfrac12 × 8 × 6 × tfrac√(3)2 = 12√(3). Multiply the whole numbers first, then attach the surd to the end of the line.
Method: 1. Write down the exact value of the sine of the angle. 2. Substitute into tfrac12absin C and cancel any halves. 3. Multiply the whole numbers together and keep the surd. 4. Leave the answer in the form asked for, with units.
Many questions give you the area and ask for a side or an angle. Substitute the numbers you know into Area = tfrac12absin C and treat it as an equation. To find a side, divide twice: for A = tfrac12absin C the missing side is b = 2A/(asin C). To find an angle, work out sin C = 2A/ab and use the inverse sine.
Be careful with the angle. The calculator gives only an acute answer, but the sine of C and the sine of 180^° - C are equal. If the question says "angle C is obtuse", subtract the calculator answer from 180^°. If it says nothing, check whether both angles are possible. Higher papers also put algebra in the sides, such as x and x + 3, which gives a quadratic that you solve and then reject any negative length.
Method: 1. Write the formula with the known numbers and the area in. 2. Rearrange for the unknown side, or for sin C. 3. Use the inverse sine for an angle, and subtract from 180^° if it is obtuse. 4. Check that your answer is a sensible length or angle.
A regular polygon with n sides splits into n identical triangles that meet at the centre. Each triangle has two sides equal to the radius and the angle at the centre is 360^° ÷ n. So the area is n × tfrac12r^2 sin((360^°)/n). A regular octagon uses eight triangles of 45^° each, and a regular pentagon uses five triangles of 72^° each.
A parallelogram with sides a and b and angle C between them has area absin C, which is two triangles. A segment is a sector minus a triangle: sector area is (θ)/360 × π r^2 and the triangle is tfrac12r^2sinθ. Calculate each piece in a separate line and subtract at the end so the method marks are easy to find.
Method: 1. Split the shape into triangles using the centre or a diagonal. 2. Find the angle in each triangle from 360^° ÷ n or from the diagram. 3. Use tfrac12absin C for one triangle, then multiply or subtract. 4. Round only at the end and write the unit.
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