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Topic 3.04 · SL and HL

Non-right angle triangles (Sine, cosine rule, area of triangle): notes and practice questions

Summary
  • Sine Rule: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
  • Cosine Rule: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C
  • Area of Triangle: Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

How it is examined

Paper 2. The choice between the two rules is the assessed decision, and the common error is using the sine rule where the given information is two sides and the included angle. Rounding is a real risk: an intermediate angle rounded to 3 significant figures and then reused loses accuracy in the final answer. 4 to 7 marks.

Given in the booklet

All four formulas are given.

Key ideas
  • Use of sine, cosine and tangent ratios to find the sides and angles of right-angled triangles.
  • The sine rule: asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}.
  • The cosine rule: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C; cos⁡C=a2+b2−c22ab\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}.
  • Area of a triangle as 12absin⁡C\dfrac{1}{2}ab\sin C.

Linking questions

  • Other contexts: triangulation, map-making.
  • International-mindedness: diagrams of Pythagoras' theorem in early Chinese and Indian manuscripts.

Practice questions

26 questions · 1 easy · 19 medium · 6 hard
Showing 20 of 20

Question 1

EasyPaper 1 · no calculator3 marks

A landscape designer is creating a triangular garden bed. Two sides of the garden bed measure 8 m and 10 m, and the angle between them is 120∘120^\circ. Find the exact area of the garden bed.

Question 2

MediumPaper 1 · no calculator6 marks

The diagram shows a parallelogram PQRS where PQ = 626\sqrt{2} cm, PS = 7 cm and cos⁡(SP^Q)=35\cos(\text{S}\hat{\text{P}}\text{Q}) = \frac{3}{5}.

Diagram of parallelogram PQRS. Side PQ is the base. Angle SPQ is indicated.

Find the exact area of the parallelogram PQRS.

Question 3

HardPaper 1 · no calculator14 marks
(a)(i)

Consider an acute angle xx such that cos⁡x=34\cos x = \frac{3}{4}.

Find the value of sin⁡x\sin x.

[2]
(a)(ii)

Find the value of cos⁡(2x)\cos(2x).

[2]
(b)

The following diagram shows triangle ABC, with BC^A=x\text{B}\hat{\text{C}}\text{A} = x, BA^C=2x\text{B}\hat{\text{A}}\text{C} = 2x, and AB = 14.

A diagram of triangle ABC. Angle at vertex C is labelled x. Angle at vertex A is labelled 2x. The side opposite vertex C, AB, is labelled 14.

(b) Show that BC = 21.

[3]
(c)(i)

The line segment CA is extended to a point D such that triangle ABD is isosceles with AB = AD.

The same triangle ABC as before. The line segment CA is extended to a point D, so that C, A, D are collinear. A line segment connects B and D, forming a new triangle ABD.

Find the size of angle ADB in terms of xx.

[3]
(c)(ii)

Find the area of triangle ABD.

[4]

Question 4

MediumPaper 1 · no calculator7 marks

The following diagram shows triangle LMN, with LM = 2152\sqrt{15}, MN = xx and LN = 3x3x.

Triangle LMN with sides 2*sqrt(15), x, and 3x. Angle LNM is at vertex N.

Given that cos⁡(∠LNM)=23\cos(\angle LNM) = \frac{2}{3}, find the area of the triangle.

Give your answer in the form aba\sqrt{b} where a,b∈Z+a, b \in \mathbb{Z}^+.

Question 5

HardPaper 1 · no calculator6 marks

In any triangle ABC, the side lengths opposite to the angles A, B, and C are a, b, and c respectively.

Show that a+bc=cos⁡(A−B2)sin⁡(C2)\frac{a+b}{c} = \frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{C}{2}\right)}.

Question 6

MediumPaper 1 · no calculator6 marks

The diagram shows a parallelogram PQRS. It is given that PQ=25PQ = 2\sqrt{5} cm, PS=8PS = 8 cm and cos⁡(QP^S)=23\cos(Q\hat{P}S) = \frac{2}{3}.

Diagram of parallelogram PQRS, not to scale

Find the area of the parallelogram PQRS.

Question 7

HardPaper 2 · calculator15 marks
(a)

A landscape architect is designing a triangular garden bed PQRPQR. The side PQPQ is fixed at a length of 6060 metres. The angle QP^RQ\hat{P}R is set to 35∘35^\circ. The architect initially plans for the side PRPR to be 4545 metres.

(a) Calculate the length of the side QRQR.

[3]
(b)

Due to a change in design, the architect now wants the angle PQ^RP\hat{Q}R to be 75∘75^\circ. The side PQPQ remains 6060 metres and QP^RQ\hat{P}R remains 35∘35^\circ.

(b) Calculate the length of the side PRPR.

[3]
(c)

(c) Calculate the area of the garden bed PQRPQR with the dimensions from part (b).

[2]
(d)

The architect considers a different design where PQ=60PQ = 60 m and QP^R=35∘Q\hat{P}R = 35^\circ (as before). However, the side QRQR is now fixed at 4040 metres.

(d) Show that two different triangular garden beds are possible, and find the two possible lengths for the side PRPR.

[7]

Question 8

MediumPaper 1 · no calculator6 marks

The following diagram shows an isosceles triangle XYZ, where XY = XZ = 5 cm and YZ = 272\sqrt{7} cm. Angle YX^\hat{\text{X}}Z = 2α2\alpha.

Diagram of an isosceles triangle XYZ with vertex X at the top. Sides XY and XZ are marked as 5cm. The base YZ is marked as 2*sqrt(7) cm. The angle at vertex X, YXZ, is marked as 2*alpha.

Find the exact value of cos⁡α\cos \alpha, giving your answer in the form p2q\frac{p\sqrt{2}}{q}, where p,q∈Z+p, q \in \mathbb{Z}^+.

Question 9

HardPaper 1 · no calculator11 marks
(a)

Find, in the form reiθre^{i\theta}, the fourth roots of −8−83i-8 - 8\sqrt{3}i.

[8]
(b)

The four roots from part (a) form the vertices of a quadrilateral in an Argand diagram. Determine the exact area of this quadrilateral.

[3]

Question 10

MediumPaper 1 · no calculator6 marks

The following diagram shows triangle PQR, with PQ = 8 cm, PR = 5\sqrt{5} cm and cos⁡(QP^R)=23\cos(\text{Q}\hat{\text{P}}\text{R}) = \frac{2}{3}.

Triangle PQR with side lengths and angle indicated

Find the area of triangle PQR.

Question 11

HardPaper 2 · calculator21 marks
(a)

Consider the non-zero vectors u⃗\vec{u} and v⃗\vec{v}. Let θ\theta be the angle between u⃗\vec{u} and v⃗\vec{v}.

Using the definitions of u⃗⋅v⃗\vec{u} \cdot \vec{v} and u⃗×v⃗\vec{u} \times \vec{v} in terms of ∣u⃗∣|\vec{u}|, ∣v⃗∣|\vec{v}| and θ\theta, show that (u⃗⋅v⃗)2+∣u⃗×v⃗∣2=∣u⃗∣2∣v⃗∣2(\vec{u} \cdot \vec{v})^2 + |\vec{u} \times \vec{v}|^2 = |\vec{u}|^2|\vec{v}|^2.

[2]
(b)(i)

A triangle PQR has vertices P(1, 0, 1), Q(a,ba, b, 2) and R(4, 1, 1), where a,b∈Qa, b \in \mathbb{Q}.

The vectors u⃗\vec{u} and v⃗\vec{v} are defined as u⃗=PQ⃗\vec{u} = \vec{PQ} and v⃗=PR⃗\vec{v} = \vec{PR}.

It is given that u⃗⋅v⃗=4\vec{u} \cdot \vec{v} = 4 and the area of triangle PQR is 142\frac{\sqrt{14}}{2} square units.

Find the value of ∣u⃗×v⃗∣|\vec{u} \times \vec{v}|.

[1]
(b)(ii)

Hence, or otherwise, find the value of ∣u⃗∣|\vec{u}|.

[4]
(b)(iii)

Hence, or otherwise, find the possible values of aa and the corresponding values of bb.

[8]
(c)

Consider a new point S, the vector w⃗\vec{w} is defined as w⃗=RS⃗\vec{w} = \vec{RS}.

It is given that u⃗⋅w⃗=0\vec{u} \cdot \vec{w} = 0 and v⃗⋅w⃗=0\vec{v} \cdot \vec{w} = 0, and the area of triangle PRS is 10 square units.

Assuming that a=2a = 2, find the possible vectors for w⃗\vec{w}.

[6]

Question 12

MediumPaper 2 · calculator5 marks

A lighthouse (L), a boat (B), and a navigation buoy (N) form a triangle.

The distance from the lighthouse to the boat, LB, is 15 km.

The distance from the lighthouse to the buoy, LN, is 9 km.

The angle at the boat, LB^N\text{L}\hat{\text{B}}\text{N}, is 30°30\degree.

Find the smallest possible perimeter of triangle LBN.

Question 13

HardPaper 1 · no calculator7 marks
(a)

A parallelogram has adjacent side lengths of 6 cm and 8 cm. Let α\alpha be the angle between these two sides. The area of the parallelogram is 12712\sqrt{7} cm2^2.

(a) Show that sin⁡α=74\sin \alpha = \frac{\sqrt{7}}{4}.

[1]
(b)

(b) Find the exact lengths of the two diagonals of the parallelogram.

[6]

Question 14

MediumPaper 2 · calculator6 marks
(a)

(a) A surveyor is mapping a new development. From point A, the distance to a prominent tree (T) is measured as 300 m. From another point B, 250 m away from A, the distance to the same tree (T) is measured as 200 m.

Find the measure of angle ATB.

[3]
(b)

(b) Find the shortest distance from the tree (T) to the line segment AB.

[3]

Question 15

MediumPaper 2 · calculator6 marks
(a)

A geological survey is being conducted in a mountainous region. Three sensor stations, A, B, and C, are set up at different locations. Their coordinates, relative to a central reference point (in meters), are given as:

Station A: (1,2,3)(1, 2, 3)

Station B: (4,5,6)(4, 5, 6)

Station C: (7,2,1)(7, 2, 1)

(a) Find the distance between Station A and Station B.

[2]
(b)

(b) Find the size of the angle ABC \text{ABC} (the angle at Station B).

[4]

Question 16

MediumPaper 2 · calculator9 marks
(a)

A landscape architect is designing a triangular shade sail for a patio. The vertices of the sail are defined by points A, B, and C in a 3D coordinate system, where the z-axis represents height.

The coordinates of the vertices are A(0,k,1)(0, k, 1), B(2,1,0)(2, 1, 0) and C(k,0,3)(k, 0, 3), where kk is a positive constant representing a design parameter.

(a) Show that the vector product AB⃗×AC⃗\vec{AB} \times \vec{AC} is given by (2−3k−k−4k2−3k)\begin{pmatrix} 2-3k \\ -k-4 \\ k^2-3k \end{pmatrix}.

[4]
(b)

(b) The architect wants to minimize the tension in the sail, which is proportional to the magnitude of the vector product of two adjacent sides. Find the smallest possible value of ∣AB⃗×AC⃗∣|\vec{AB} \times \vec{AC}|.

[3]
(c)

(c) Calculate the smallest possible area of the shade sail.

[2]

Question 17

MediumPaper 2 · calculator8 marks

A lighthouse (L) monitors maritime traffic. At a particular moment, the lighthouse observes a buoy (B) and a ship (S).

The bearing of the buoy from the lighthouse is 050∘050^\circ and its distance is 1515 km. The bearing of the ship from the lighthouse is 170∘170^\circ and its distance is 2525 km.

Calculate the distance and bearing of the buoy (B) from the ship (S).

Question 18

MediumPaper 1 · no calculator6 marks
(a)

A plan for a garden is drawn on a coordinate grid, where 1 unit represents 1 metre. The garden consists of a paved patio and a flower bed.

The patio is a quadrilateral with vertices A(-4, 2), B(0, 5), C(4, 2), and D(0, -1).

(a) Find the area of the patio ABCD.

[3]
(b)

The flower bed is a triangle with vertices C(4, 2), E(9, 4), and F(9, 0).

(b) Find the area of the flower bed CEF.

[2]
(c)

(c) Hence, find the total area of the garden.

[1]

Question 19

MediumPaper 2 · calculator17 marks
(a)

(a) Calculate the shortest distance from the apex VV to the side ABAB of the pyramid's base.

[3]
(b)

(b) Find the total surface area of the four triangular glass panels forming the pyramid's roof.

[4]
(c)

(c) Calculate the total height of the monument from the ground to the apex VV.

[4]
(d)

(d) Determine the size of the angle between the slant edge VAVA and the base edge ABAB.

[3]
(e)

(e) A security camera is installed at a point SS on the vertical edge C′CC'C of the monument. A guard standing at point B′B' (a corner of the base of the monument) measures the angle of elevation to point SS to be 40∘40^\circ. Find the height of the security camera from the ground.

[3]

Question 20

MediumPaper 2 · calculator8 marks
(a)

(a) A landscape architect is designing a new park. From a central monument, two paths, Path A and Path B, diverge. Path A is 7575 m long, and Path B is 6060 m long. The angle between Path A and Path B at the monument is 125∘125^\circ.

Calculate the direct distance between the ends of Path A and Path B.

[3]
(b)

(b) Calculate the area of the triangular region enclosed by Path A, Path B, and the direct line connecting their ends.

[2]
(c)

(c) Calculate the angle that Path B makes with the direct line connecting the ends of the paths.

[3]

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What does Non-right angle triangles (Sine, cosine rule, area of triangle) cover in IB Maths AA?

Sine Rule: (a)/(sin A) = (b)/(sin B) = (c)/(sin C). Cosine Rule: c^2 = a^2 + b^2 - 2abcos C. Area of Triangle: Area = (1)/(2)absin C.

Is Non-right angle triangles (Sine, cosine rule, area of triangle) SL or HL?

Both. SL and HL students study Non-right angle triangles (Sine, cosine rule, area of triangle) to the same depth.

How do I revise Non-right angle triangles (Sine, cosine rule, area of triangle) for IB Maths AA?

Start from the core idea: sine Rule: (a)/(sin A) = (b)/(sin B) = (c)/(sin C). In the exam: paper 2. The choice between the two rules is the assessed decision, and the common error is using the sine rule where the given information is two sides and the included angle. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Non-right angle triangles (Sine, cosine rule, area of triangle)?

FourtyFive has 26 Non-right angle triangles (Sine, cosine rule, area of triangle) questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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