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secondary 4 | A Maths
3 Answers Below
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cos θ = -1/5
(-ve , θ in 2nd or 3rd quad)
cos θ = -1/5
this means
adj = -1
hyp = 5
opp = √[ 5² - (-1)² ] = √24
so
sin θ = opp/hyp = √24/5
(a)
sin [ cos⁻¹ (-1/5) ]
= sin θ
= √24/5
let angle θ = sin⁻¹ (-2/3)
sin θ = -2/3
(-ve , θ in 3rd or 4th quad)
sin θ = -2/3
this means
opp = -2
hyp = 3
adj = √[ 3² - (-2)² ] = √5
so
tan θ = opp/adj = -2/√5
(b)
tan [ sin⁻¹ (-2/3) ]
= tan θ
= -2/√5
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