Year 9 maths · Chapter 3

Year 9 Pythagoras and trigonometry: revision mind map

Pythagoras' theorem in 2D and 3D, the three trig ratios (SOH CAH TOA), finding sides and angles, elevation, depression and bearings. Everything on one A4 page.

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1Pythagoras' theorem

In a right-angled triangle: c2=a2+b2c^2 = a^2 + b^2. cc is the hypotenuse, the longest side.

Sides 6 and 8: c2=36+64=100c^2 = 36 + 64 = 100, so c=10c = 10

abc (hypotenuse)

2Finding a shorter side

To find a shorter side, subtract: a2=c2−b2a^2 = c^2 - b^2.

Hypotenuse 13, side 5: a2=169−25=144a^2 = 169 - 25 = 144, so a=12a = 12

Watch out: a shorter side can never be longer than the hypotenuse.

3Pythagoras in 2D problems

Draw it, mark the right angle, then find the triangle.

A 5 m ladder reaches 4 m up a wall: the foot is 25−16=3\sqrt{25 - 16} = 3 m out

Diagonal of a 9 by 12 rectangle: 81+144=15\sqrt{81 + 144} = 15

4Pythagoras in 3D

Find a right triangle on one face first, then use that length in a second triangle.

Box 3 by 4 by 12: base diagonal 9+16=5\sqrt{9 + 16} = 5

Space diagonal 52+122=13\sqrt{5^2 + 12^2} = 13

Shortcut: d=a2+b2+c2d = \sqrt{a^2 + b^2 + c^2}

5Trigonometric ratios

SOH CAH TOA. Label the sides from the angle you're using.

sin⁡θ=OH\sin\theta = \dfrac{O}{H}   cos⁡θ=AH\cos\theta = \dfrac{A}{H}   tan⁡θ=OA\tan\theta = \dfrac{O}{A}

θadjacent (A)opposite (O)hypotenuse (H)

Watch out: set your calculator to degrees.

6Finding unknown sides

Pick the ratio that uses the side you know and the side you want.

Opposite 35°, hypotenuse 10: x=10sin⁡35∘≈5.74x = 10\sin 35^\circ \approx 5.74

Adjacent 8, angle 40°, find opposite: x=8tan⁡40∘≈6.71x = 8\tan 40^\circ \approx 6.71

7Solving for the denominator

When xx is on the bottom, swap it with the trig value.

cos⁡30∘=12x⇒x=12cos⁡30∘≈13.86\cos 30^\circ = \dfrac{12}{x} \Rightarrow x = \dfrac{12}{\cos 30^\circ} \approx 13.86

Watch out: check the answer makes sense: the hypotenuse must be the longest side.

8Finding unknown angles

Use the inverse buttons: sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, tan⁡−1\tan^{-1}.

Opposite 5, hypotenuse 10: sin⁡θ=0.5\sin\theta = 0.5, so θ=30∘\theta = 30^\circ

Opposite 7, adjacent 4: θ=tan⁡−1(1.75)≈60.3∘\theta = \tan^{-1}(1.75) \approx 60.3^\circ

9Applying trigonometry

Angles of elevation (looking up) and depression (looking down) are measured from the horizontal.

20 m from a tree, the angle of elevation to the top is 32°

Height =20tan⁡32∘≈12.5= 20\tan 32^\circ \approx 12.5 m

Watch out: draw the horizontal line first, then the angle.

10Bearings

True bearings are measured clockwise from north, with three digits.

North-east is 045°T, south-east 135°T, west 270°T

Walk 10 km on 060°T: east 10sin⁡60∘≈8.6610\sin 60^\circ \approx 8.66 km, north 10cos⁡60∘=510\cos 60^\circ = 5 km

Year 9 maths · Chapter 3 · Mind map

Pythagoras and trigonometry

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In this chapter

  • 3A Pythagoras' theorem
  • 3B Finding a shorter side
  • 3C Pythagoras in 2D problems
  • 3D Pythagoras in 3D
  • 3E Trigonometric ratios
  • 3F Finding unknown sides
  • 3G Solving for the denominator
  • 3H Finding unknown angles
  • 3I Applying trigonometry
  • 3J Bearings

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