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Topic 3.03 · SL and HL

Right angle triangles (SOH CAH TOA): notes and practice questions

Summary
  • Pythagorean Theorem is used only for right-angled triangles. It states that the square of the hypotenuse (cc) equals the sum of the squares of the two shorter sides (aa and bb): a2+b2=c2a^2 + b^2 = c^2. You must remember this formula as it is not provided in the booklet.
  • Trigonometry (SOHCAHTOA) relates the ratios of side lengths to an angle (θ\theta) in a right-angled triangle.
  • SOH: Sin θ=OppositeHypotenuse\text{Sin } \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}
  • CAH: Cos θ=AdjacentHypotenuse\text{Cos } \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}
  • TOA: Tan θ=OppositeAdjacent\text{Tan } \theta = \frac{\text{Opposite}}{\text{Adjacent}}.
  • You can use these ratios with inverse functions (like Sin−1\text{Sin}^{-1}) to find missing angles, or directly to find missing side lengths, as long as you label the sides correctly relative to the angle θ\theta.

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

11 questions · 10 medium · 1 hard
Showing 11 of 11

Question 1

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 2

HardPaper 1 · no calculator19 marks
(a)

A ladder must be placed against a tall vertical building, clearing a monument that is 8 m high and stands on horizontal ground 1 m away from the building's base. The ladder touches the ground, the top corner of the monument, and the wall of the building.

A diagram showing a vertical building and the horizontal ground. A monument of height 8m stands 1m away from the base of the building. A ladder is shown leaning against the building, just touching the top of the monument. The angle the ladder makes with the ground is labelled as theta.

Let LL be the length of the ladder in metres.

Let θ\theta be the angle that the ladder makes with the ground, where 0<θ<π20 < \theta < \frac{\pi}{2}.

(a) Show that L=sec⁡θ+8csc⁡θL = \sec \theta + 8\csc \theta.

[2]
(b)(i)

(b) (i) Find dLdθ\frac{dL}{d\theta}.

[2]
(b)(ii)

(b) (ii) When dLdθ=0\frac{dL}{d\theta} = 0, show that tan⁡θ=2\tan\theta = 2.

[3]
(c)(i)

(c) (i) Find d2Ldθ2\frac{d^2L}{d\theta^2}.

[3]
(c)(ii)

(c) (ii) When tan⁡θ=2\tan\theta = 2, find the value of d2Ldθ2\frac{d^2L}{d\theta^2}.

[4]
(d)(i)

(d) (i) Hence, justify that LL is a minimum when tan⁡θ=2\tan\theta = 2.

[1]
(d)(ii)

(d) (ii) Determine this minimum value of LL.

[2]
(e)

(e) A construction company only has ladders with a maximum length of 11 m. Determine whether it is possible to position a ladder against the building over the monument, giving a reason for your answer.

[2]

Question 3

MediumPaper 2 · calculator15 marks
(a)

(a) A new Ferris wheel is being constructed. From a point A on the ground, 15 metres from the base of the support tower, the angle of elevation to the centre C of the Ferris wheel is 33.7 degrees. Find the height of point C above the ground.

[2]
(b)

(b) An engineer walks 5 metres closer to the support tower to point B. Find the angle of elevation of point C from point B, giving your answer in radians.

[2]
(c)

(c) The lowest point a passenger cabin reaches is 1.5 metres above the ground, allowing for easy boarding. Calculate the radius of the Ferris wheel.

[2]
(d)

(d) The height hh, in metres, of a passenger cabin above the ground can be modelled by the function h(t)=q−pcos⁡(kt)h(t) = q - p \cos(kt), where tt is the time in seconds after the cabin starts moving from its lowest point. The Ferris wheel completes one full rotation in 40 seconds.

Find the values of pp, qq, and kk.

[6]
(e)

(e) An observer watches the Ferris wheel for 5 minutes. How many times does a specific cabin pass its highest point during this time?

[3]

Question 4

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 5

MediumPaper 1 · no calculator5 marks

A decorative object is in the shape of a right square-based pyramid. The base of the pyramid is a square with side length 10 cm10 \text{ cm}. The perpendicular height of the pyramid is 12 cm12 \text{ cm}.

Find the total surface area of the object.

Question 6

MediumPaper 2 · calculator6 marks

From the top of a lighthouse 90 m90 \text{ m} high, an observer spots a boat at sea. The angle of depression to the boat is 55∘55^\circ. The boat sails directly away from the lighthouse, and after eight minutes the angle of depression is 15∘15^\circ. Calculate the speed of the boat in kmh−1\text{kmh}^{-1}.

Question 7

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 8

MediumPaper 2 · calculator9 marks
(a)

(a) A surveyor is measuring a tall communications mast. From a point PP on the ground, the angle of elevation to the top of the mast is 38∘38^\circ. The total height of the mast from the ground to its top is 7070 meters. Calculate the horizontal distance dd from point PP to the near edge of the mast's base, giving your answer correct to one decimal place.

[3]
(b)

(b) Calculate the direct line-of-sight distance LL from point PP to the top of the mast, giving your answer correct to one decimal place.

[3]
(c)

(c) A maintenance platform is located 1515 m below the top of the mast. Find the angle of depression from this maintenance platform to point PP, giving your answer in degrees correct to one decimal place.

[3]

Question 9

MediumPaper 2 · calculator11 marks
(a)

A designer is creating a symmetrical display stand for a museum exhibit. The cross-section of the stand is an isosceles trapezium ABCDABCD, where BCBC is parallel to ADAD. The shorter parallel side BCBC measures 1515 cm, the longer parallel side ADAD measures 2727 cm, and the non-parallel sides ABAB and CDCD each measure 1010 cm.

Show that the height of the trapezium is 88 cm.

[3]
(b)

Hence, find the area of the cross-section of the display stand.

[2]
(c)

Find the size of the angle ADCADC, giving your answer to one decimal place.

[3]
(d)

Calculate the length of the diagonal ACAC, giving your answer to three significant figures.

[3]

Question 10

MediumPaper 2 · calculator6 marks
(a)

(a) The main antenna of a communication tower is located at point A(2,8,10)A(2, 8, 10). The centre of the rectangular base of the tower is at point C(0,5,0)C(0, 5, 0). A vertical support beam connects the antenna to the centre of the base. Calculate the length of this support beam, ACAC.

[2]
(b)

(b) The rectangular base of the tower has dimensions 66 m by 44 m. Calculate the length of the diagonal of this base.

[2]
(c)

(c) A support cable runs from the antenna AA to one of the corners of the base, say point PP. Find the size of the angle that this support cable APAP makes with the base platform.

[2]

Question 11

MediumPaper 2 · calculator15 marks
(a)

A company produces solid glass paperweights. One model is a right pyramid with a square base of side length 10 cm and a height of 12 cm.

Find the volume of one of these paperweights.

[2]
(b)

Find the slant height of the pyramid.

[2]
(c)

The company is investigating a new design for a paperweight, a cuboid with the same square base and the same total surface area as the pyramid.

Show that the total surface area of the pyramid is 360 cm2^2.

[3]
(d)

Find the height, HH, of the cuboid-shaped paperweight.

[4]
(e)

The company wants to choose the design that uses less glass in order to reduce production costs.

State whether they should switch to the cuboid design. Justify your conclusion.

[4]

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What does Right angle triangles (SOH CAH TOA) cover in IB Maths AA?

Pythagorean Theorem is used only for right-angled triangles. It states that the square of the hypotenuse (c) equals the sum of the squares of the two shorter sides (a and b): a^2 + b^2 = c^2. You must remember this formula as it is not provided in the booklet. Trigonometry (SOHCAHTOA) relates the ratios of side lengths to an angle (θ) in a right-angled triangle. SOH: Sin θ = fracOppositeHypotenuse.

Is Right angle triangles (SOH CAH TOA) SL or HL?

Both. SL and HL students study Right angle triangles (SOH CAH TOA) to the same depth.

How do I revise Right angle triangles (SOH CAH TOA) for IB Maths AA?

Start from the core idea: pythagorean Theorem is used only for right-angled triangles. It states that the square of the hypotenuse (c) equals the sum of the squares of the two shorter sides (a and b): a^2 + b^2 = c^2. You must remember this formula as it is not provided in the booklet. 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 Right angle triangles (SOH CAH TOA)?

FourtyFive has 11 Right angle triangles (SOH CAH TOA) 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.

Is FourtyFive free for Right angle triangles (SOH CAH TOA) practice?

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Can I handwrite Right angle triangles (SOH CAH TOA) answers on an iPad?

Yes. In the FourtyFive iPad app you write your working by hand with Apple Pencil, the way you would on paper, and it is marked the same way.

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