Sine Rule, Cosine Rule and Triangle Area
When a triangle has no right angle, SOH CAH TOA alone is not enough. The sine rule, cosine rule and trigonometric area formula extend your toolkit to general triangles.
Higher. Build on right-angled trigonometry, exact values and rearranging formulae.
Label opposite pairs and choose a method
In triangle ABC, side a is opposite angle A, b is opposite B and c is opposite C. These rules work in triangles without a right angle. Draw a sketch and match each side to its opposite angle before substituting.
Sine rule: use when you know a matching side–opposite-angle pair and one more side or angle. Cosine rule: use for two sides and their included angle, or three sides. Area: use two sides and their included angle.
Sine rule — finding a side
a/sin A = b/sin B = c/sin C. Use only the two fractions you need. It follows by drawing a perpendicular height: h=b sin A=a sin B, then rearranging.
Worked example
A=40°, B=65° and a=8 cm. Then b/sin65°=8/sin40°, so b=8sin65°/sin40°≈11.3 cm (3 significant figures). The larger angle has the longer opposite side, which checks the result.
Sine rule — finding an angle
Use the reciprocal form sin A/a=sin B/b, rearrange for the sine of the missing angle, then use inverse sine.
Worked example — an acute angle specified
A=40°, a=8 cm, b=10 cm, and B is stated to be acute. sinB=10sin40°/8≈0.8034845, so B≈53.5°.
Check ambiguity: sinθ=sin(180°−θ). If B were not stated acute, B≈126.5° would also fit here: both 40°+53.5° and 40°+126.5° are below 180°. The remaining angles are approximately 86.5° and 13.5°. Use the diagram’s stated conditions, not its apparent shape, to select a solution. A sine result greater than 1 indicates impossible data or an error.
Cosine rule — finding a side
a²=b²+c²−2bc cos A. A must be the angle between b and c.
Worked example
Two sides are 7 cm and 9 cm with included angle 60°. Opposite side a satisfies a²=49+81−2×7×9×½=67. Therefore a=√67 cm≈8.19 cm.
At 90°, cos90°=0 and the cosine rule becomes Pythagoras. The extra term adjusts for triangles that are not right-angled.
Cosine rule — finding an angle
Rearrange: cos A=(b²+c²−a²)/(2bc). Then apply cos⁻¹. Keep the numerator in brackets on your calculator.
Worked example — obtuse angle
The sides are a=9 cm, b=5 cm, c=6 cm. cosA=(25+36−81)/(2×5×6)=−1/3, so A≈109.5°. A negative cosine gives an obtuse triangle angle, consistent with A being opposite the longest side.
Triangle area and reverse problems
Area=½ab sin C, where C is the included angle between a and b. This comes from area=½×base×perpendicular height and height=b sin C.
Worked example — exact area
Sides 8 cm and 10 cm enclose 30°. Area=½×8×10×sin30°=20 cm².
Worked example — recover an angle
Area=12 cm² and sides 6 cm and 8 cm enclose angle C. Then 12=24sinC, so sinC=½. Thus C=30° or 150°; if C is specified obtuse, choose 150°.
Worked example — segment of a circle
A circle has radius 6 cm and a minor sector angle 60°. Sector area=(60/360)π×36=6π cm². The triangle formed by its radii has area ½×6×6×sin60°=9√3 cm². Minor segment area=6π−9√3 cm².
Bearings application
B is 8 km due east of A. C is on bearing 040° from A and 340° from B. Draw north lines. At A, the interior angle is 90°−40°=50°. At B, BA has bearing 270°, so the interior angle is 340°−270°=70°. Angle C=60°. Then AC/sin70°=8/sin60°, giving AC≈8.68 km.
Think Like a Mathematician
Choose the rule from the information available, not from the shape of the sketch. Ask whether you have an opposite pair or an included angle, and check that the longest side faces the largest angle.
Common Mistakes
Match side a with angle A. Use the included angle in the cosine and area rules. Take a square root after finding a². Check a second possible sine-rule angle when appropriate. Keep degree mode and round once at the end.
Practice — non-right-angled triangles
1. A=30°, B=45° and a=6 cm. Find b exactly.
b=6sin45°/sin30°=6√2 cm.
2. Two sides 5 cm and 8 cm enclose 60°. Find the third side.
a²=25+64−2×5×8×½=49, so a=7 cm.
3. Sides 4 cm, 5 cm and 7 cm form a triangle. Find the angle opposite 7 cm to 1 decimal place.
cosA=(16+25−49)/40=−0.2, so A≈101.5°.
4. Find the exact area between sides 6 cm and 10 cm with included angle 60°.
½×6×10×√3/2=15√3 cm².
5. sinB=0.6. State the two possible triangle angles to 1 decimal place.
36.9° and 143.1°, subject to the other angles and conditions.
6. A triangle has area 15 cm² and an included angle 30° between sides 10 cm and x cm. Find x.
15=½×10×x×½=2.5x, so x=6 cm.
Frequently Asked Questions
Does the sine rule always give two triangles?
No. A second candidate may violate the angle sum or other conditions. Two sides and a non-included angle can sometimes give two solutions.
Can I use the cosine rule on a right triangle?
Yes, but it reduces to Pythagoras and is usually a longer calculation.