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secondary 4 | E Maths
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Date Posted: 3 years ago
Views: 247
sin 20° = p
opp/hyp = p
p(hyp) = opp
Using Pythagoras' Theorem,
(hyp)² = (opp)² + (adj)²
(hyp)² = p²(hyp)² + (adj)²
(hyp)² - p² (hyp)² = (adj)²
(1 - p²) (hyp)² = (adj)²
1 - p² = (adj)² / (hyp)²
√(1 - p)² = adj/hyp = cos 20°
opp/hyp = p
p(hyp) = opp
Using Pythagoras' Theorem,
(hyp)² = (opp)² + (adj)²
(hyp)² = p²(hyp)² + (adj)²
(hyp)² - p² (hyp)² = (adj)²
(1 - p²) (hyp)² = (adj)²
1 - p² = (adj)² / (hyp)²
√(1 - p)² = adj/hyp = cos 20°
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done
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hope this helps (-:
Date Posted:
3 years ago
Oh thx
Things to note :
The opposite isn't necessarily p and the hypotenuse isn't necessarily 1.
The ratio of opposite to hypotenuse however, is p : 1
So we could also have possibilities like :
Opposite = 2p, hypotenuse = 2
Opposite = p/6, hypotenuse = 1/6
Opposite = 53⅓ p, hypotenuse = 53⅓
The opposite isn't necessarily p and the hypotenuse isn't necessarily 1.
The ratio of opposite to hypotenuse however, is p : 1
So we could also have possibilities like :
Opposite = 2p, hypotenuse = 2
Opposite = p/6, hypotenuse = 1/6
Opposite = 53⅓ p, hypotenuse = 53⅓
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