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secondary 4 | E Maths
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please help! thanks
Date Posted: 2 years ago
Views: 309
Q(3)
cos θ = ⁵⁄₁₃ = b/c
b = 5
c = 13
a² = c² - b²
= 13² - 5²
= 144
= 12²
a = 12
cosec θ = 1/ ( sin θ )
= c/a
= ¹³⁄₁₂ (Ans)
Q(4)
b = k
a = -1
c = √b²+a²
= √k²+1
sec A = 1/ (cos A)
= c/k
= (√k²+1) / k (Ans)
cos θ = ⁵⁄₁₃ = b/c
b = 5
c = 13
a² = c² - b²
= 13² - 5²
= 144
= 12²
a = 12
cosec θ = 1/ ( sin θ )
= c/a
= ¹³⁄₁₂ (Ans)
Q(4)
b = k
a = -1
c = √b²+a²
= √k²+1
sec A = 1/ (cos A)
= c/k
= (√k²+1) / k (Ans)
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Date Posted:
2 years ago
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