1. Learning how to … use Pythagoras’ Theorem to calculate a shorter side of a right-angled triangle Mathematics GCSE Topic Reminders

2. We already know how to … identify the hypotenuse … Hypotenuse Opposite the right angle

3. We already know … Pythagoras ’ Theorem … += For a right-angled triangle … Hypotenuse Biggest square …on its own

4. We already know … Pythagoras ’ Theorem … += For a right-angled triangle … Biggest square a b c a2a2 b2b2 c2c2 a2a2 b2b2 c2c2

5. 3, 4, 5 example 16 + 9 = 25

6. We already know … Pythagoras ’ Theorem … += Biggest square a b c We don’t need to keep drawing the squares … a2a2 b2b2 c2c2 a2a2 b2b2 c2c2

7. We already know how to … square a number … 7272 means … 7 x 7= 49 4.8 2 on a calculator … 4.8 x 4.8 or … NOT 7 x 2

8. 23.04

9. We already know how to … find the square root of a number …  36 means … ? x itself = 36 i.e. 6  52 on a calculator … because 6 x 6 = 36

10. 23.04 7.2 to 1dp 7.21110255

11. We already know how to … calculate the hypotenuse … += Biggest square 4.7 6.2 c 4.7 2 6.2 2 c2c2 4.7 2 6.2 2 c2c2 + =22.0938.44c2c2 =60.53c2c2 =  60.53c =7.8c

12. We will now learn how to … calculate a shorter side … Shorter side Hypotenuse

13. Identify the hypotenuse … opposite the right angle Label the unknown side x

14. Write down the square of each side Write Pythagoras’ Rule using these squares x x2x2 16.5 2 12.8 2 Biggest square x2x2 + 12.8 2 = 16.5 2 GCSE 1 st

15. Rearrange to leave x 2 on its own x x2x2 +12.8 2 =16.5 2 x 2 = 16.5 2 - 12.8 2 x2x2 -163.84=272.25 Calculate the right hand side to give x 2 x 2 = 108.41 GCSE 2 nd

16. Find the  of this to give x x 2 = 108.41 x =  108.41 Write answer to appropriate degree of accuracy and refer to original question x = 10.41201 radius = 10.4 cm x GCSE 3 rd

17. Calculate the length of AC, giving your answer to a suitable degree of accuracy. Calculate the length of AD, giving your answer to a suitable degree of accuracy.

18. PQRS is a parallelogram with SR = PQ = 15.6 and PS = QR = 9.8cm. M is the foot of the perpendicular from P onto SR and SM = 4.7cm. Find the length of PM.

19. Use Pythagoras’ Theorem to calculate the height marked h.

20. Learning how to … use trigonometry to find an angle in a right angled triangle. Mathematics GCSE Topic Reminders

21. We already know how to … label the sides … x Opposite Hypotenuse Opposite the right angle Adjacent Opposite the angle we need Next to the angle we need

22. We already know … The formula for SINE … x Opposite Hypotenuse Sin x = Opposite Hypotenuse Adjacent

23. We already know … The formula for COSINE … x Opposite Hypotenuse Cos x = Adjacent Hypotenuse Adjacent

24. We already know … The formula for TANGENT … x Opposite Hypotenuse Tan x = Opposite Adjacent

25. Don’t use a protractor

26. Label the given sides in relation to the angle you need to find Opposite Hypotenuse Adjacent

27. Opposite Choose SIN, COS or TAN depending on the sides given ‘Opposite’ & ‘Adjacent’ means we’ll be using TAN for this question Adjacent

28. Opposite Adjacent Write out the formula with correct numbers Tan x = 138opposite 177adjacent GCSE 1 st

29. Opposite Adjacent Tan x = 138 177 Key 2 nd (or INV ) TAN 138  177 = into the calculator to give the angle x Calculator set to DEG x = 37.9 o GCSE 2 nd & 3 rd

31. Calculate the size of BĎE.

32. Calculate the length of BE.