Area of a Triangle (AI SL)
When you know two sides of a triangle and the angle trapped between them, you can find its area without ever measuring a height. This sub-topic covers that formula in both directions - using it to find an area, and rearranging it to find a missing angle or side. It's part of the broader Triangle Trigonometry topic.
26 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference SL3.2, which includes the area of a triangle as \(\tfrac12 ab\sin C\). It's in the formula booklet in its standard form; rearranging it to solve for an angle is a skill you supply yourself.
Area from two sides and the included angle
\(\text{Area}=\tfrac12 ab\sin C\)
\(a\) and \(b\) are two sides, \(C\) is the angle trapped between them.
✓ In the formula bookletRearranged - find the angle
\(\sin C=\dfrac{2\times\text{Area}}{ab}\)
Given the area and two sides, solve for \(C\) - remember \(\sin^{-1}\) can give an acute or an obtuse answer.
Not itself in the booklet - rearrange the area formulaNeed the wider syllabus context and GDC settings for trig? See Triangle Trigonometry's GDC guidance.
Worked examples
Find the area of a triangle with sides 10 cm and 14 cm and included angle \(40^\circ\) (3 significant figures).
Worked solution
Two sides and the included angle give \(A = \tfrac12 ab\sin C\). M1
\(a=10,\ b=14,\ C=40^\circ\): \(A = \tfrac12(10)(14)\sin 40^\circ = 70\sin 40^\circ = 70(0.6428\ldots) = 44.99\ldots\) A1
\(A \approx 45.0 \text{ cm}^2\) (3 significant figures). A1
A triangular plot has two sides of 25 m and 30 m. The area is \(300\) m\(^2\).
(a)(i) Find the value of \(\theta<90\).
(a)(ii) Find the value of \(\theta>90\).
(b)(i) Find the perimeter for the smaller-angle case.
(b)(ii) Find the perimeter for the larger-angle case.
Worked solution
(a) \(\tfrac12(25)(30)\sin\theta=300\Rightarrow\sin\theta=\dfrac{600}{750}=0.8.\) M1 A1
\(\theta_1=53.1^\circ\), A1 \(\theta_2=126.9^\circ.\) A1
(b) For \(\theta_1=53.1^\circ\) (exactly \(\cos\theta_1=0.6\)): \(c^2=625+900-1500(0.6)=625\Rightarrow c=25.0\) m; perimeter \(=25+30+25.0=80.0\) m. M1 A1
For \(\theta_2=126.9^\circ\) (\(\cos\theta_2=-0.6\)): \(c^2=625+900-1500(-0.6)=2425\Rightarrow c\approx49.2\) m M1 ; perimeter \(\approx25+30+49.2=104.2\) m. A1
Common mistakes
- Using the wrong pair of sides. The two lengths in \(\tfrac12 ab\sin C\) must be the sides either side of the angle \(C\), not any two sides of the triangle.
- Dropping the second solution when solving for an angle. Rearranging to \(\sin C = k\) gives both an acute and an obtuse angle - read the question carefully to see whether it wants one, the other, or both.
- Forgetting units are squared. Area answers need cm\(^2\), m\(^2\) etc, not the linear unit used for the sides.
Ready to practise properly?
26 triangle-area questions, marked instantly like the real exam.
Quick answers
What is the formula for the area of a triangle using trigonometry?
\(\text{Area} = \tfrac12 ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the angle trapped between them. It works for any triangle, not just right-angled ones.
Can the area formula give two possible angles?
Yes. Solving \(\sin C = k\) for \(C\) gives an acute solution and a supplementary obtuse solution (\(180^\circ\) minus the acute value), since sine is positive in both the first and second quadrants - check whether the question wants one, or both.