Area using ½ab sin C questions with answers

Questions

1. Question 1Find the area: acute angle

   In triangle $ABC$, $BC = 7$ cm, $CA = 8$ cm and angle $ACB = 57^\circ$. Find the area of the triangle, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Two sides and the angle between them$\text{Area} = \dfrac{1}{2}ab \sin C$$a = 7$, $b = 8$, $C = 57^\circ$
   2. Step 2: Substitute and work it out$\text{Area} = \dfrac{1}{2} \times 7 \times 8 \times \sin 57^\circ$$\text{Area} = 23.4827\ldots$$\phantom{\text{Area}} = 23.5$ cm² (3 s.f.)
2. Question 2Find the area: obtuse angle

   In triangle $ABC$, $BC = 7$ cm, $CA = 10$ cm and angle $ACB = 127^\circ$. Find the area of the triangle, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Two sides and the angle between them$\text{Area} = \dfrac{1}{2}ab \sin C$$a = 7$, $b = 10$, $C = 127^\circ$
   2. Step 2: Substitute and work it out$\text{Area} = \dfrac{1}{2} \times 7 \times 10 \times \sin 127^\circ$$\text{Area} = 27.9522\ldots$$\phantom{\text{Area}} = 28.0$ cm² (3 s.f.)
3. Question 3Find the angle from the area

   In triangle $ABC$, $BC = 5$ cm, $CA = 14$ cm and the area is $25$ cm². Given that angle $ACB$ is acute, find angle $ACB$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Substitute into the area formula$\text{Area} = \dfrac{1}{2}ab \sin C$$25 = \dfrac{1}{2} \times 5 \times 14 \times \sin \theta$
   2. Step 2: Make sin θ the subject$\sin \theta = \dfrac{2 \times 25}{5 \times 14}$$\phantom{\sin \theta} = 0.7142\ldots$
   3. Step 3: Use sin⁻¹$\theta = \sin^{-1}(0.7142\ldots)$$\theta = 45.5846\ldots^\circ$$\phantom{\theta} = 45.6^\circ$ (1 d.p.)
4. Question 4Find the obtuse angle from the area

   In triangle $ABC$, $BC = 11$ cm, $CA = 12$ cm and the area is $40$ cm². Given that angle $ACB$ is obtuse, find angle $ACB$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Substitute into the area formula$\text{Area} = \dfrac{1}{2}ab \sin C$$40 = \dfrac{1}{2} \times 11 \times 12 \times \sin \theta$
   2. Step 2: Make sin θ the subject$\sin \theta = \dfrac{2 \times 40}{11 \times 12}$$\phantom{\sin \theta} = 0.6060\ldots$
   3. Step 3: The calculator gives the acute angle; θ is obtuse, so use sin θ = sin(180° − θ)Acute angle $= \sin^{-1}(0.6060\ldots)$$= 37.3052\ldots^\circ$$\sin \theta = \sin(180^\circ - 37.3052\ldots^\circ)$$\theta = 180^\circ - 37.3052\ldots^\circ$$\phantom{\theta} = 142.7^\circ$ (1 d.p.)
5. Question 5Find a side from the area

   In triangle $ABC$, $BC = 7$ cm, angle $ACB = 48^\circ$ and the area is $36$ cm². Find the length of $CA$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Substitute into the area formula$\text{Area} = \dfrac{1}{2}ab \sin C$$36 = \dfrac{1}{2} \times 7 \times x \times \sin 48^\circ$
   2. Step 2: Make x the subject$x = \dfrac{2 \times 36}{7 \sin 48^\circ}$$x = 13.8407\ldots$$\phantom{x} = 13.8$ cm (3 s.f.)