FRIDAY LECTURE WORKED SOLUTIONS.pdf

FRIDAY LECTURE WORKED SOLUTIONS.pdf - FRIDAY LECTURE WORKED...

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Unformatted text preview: FRIDAY LECTURE WORKED SOLUTIONS Final Pythagoras Rule Questions 10 cm In this example the perpendicular height, the radius and the slant side form a right-­‐angled triangle. We will label the radius as side b. ! + ! = ! ! + 4! = 10! ! + 16 = 100 ! = 100 − 16 ! = 84 = 84 = 9.2 cm 4 cm 1.3 m 2 m Here the strut, the wall and the bottom of the sign form a right-­‐angled triangle. The distance down the wall is the b side. ! + ! = ! ! + 1.3! = 2! ! + 1.69 = 4 ! = 4 − 1.69 ! = 2.31 = 2.31 = 1.5 m 60 cm 80 cm ! + ! = ! ! + 60! = 80! ! + 3600 = 6400 ! = 6400 − 3600 ! = 2800 = 2800 = 52.9 Here the diagonal, side and top of the screen form a right-­‐angled triangle. The width of the screen is the b side. 12 cm 9 cm In this question the perpendicular height, the radius and the slant side form a right-­‐angled triangle. We will label the perpendicular height as side b. ! + ! = ! ! + 9! = 12! ! + 81 = 144 ! = 144 − 81 ! = 63 = 63 = 7.9 cm Final Trigonometric Ratio Questions Plane landing – this is a TAN question as we know the adjacent side and are finding the opposite side x cm (O) ! 3 40 km (A) = 40 40×3! = 40×0.0524 = 2.1 = The side marked x is 2.1 km Kite board rider question – this is a SIN question as we know the hypotenuse and are finding the opposite side. 80 cm (H) x cm (O) 50! = ℎ 3! = 80 80×50! = 80×0.766 = 375.5 = The side marked x is 375.5m Water ski jump question – this is a COS question as we know the hypotenuse and are finding the adjacent side. 6.8 m (H) ! 15 x m (A) = ℎ 15! = 6.8 6.8×15! = 6.8×0.9659 = 6.57 = The side marked x is 6.57 m 50! =...
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