Using the measurements given in the figure alongside,
(a) Find the values of:
(i) sin ϕ
(ii) tan θ.
(b) Write an expression for AD in terms of θ.
Answer :
BC = 12, BD = 13
In right angled ∆BCD
BD2 = BC2 + CD2
CD2 = BD2 – BC2
CD2 = (13)2 – (12)2
⇒ CD2 = 169 – 144 = 25
CD = √25 = 5
CD = BE = 5 and EA = AE = 14 – 5 = 9
(a)
(i) sin ϕ = perpendicular/hypotenuse
In right angled ∆BCD
sin ϕ = CD/BD = 5/13
(ii) tan θ = perpendicular/hypotenuse
In right angled ∆AED
tan θ = ED/AE = BC/AE = 12/9 = 4/3 (Since ED = BC)
(b) In right angled ∆AED
sin θ = perpendicular/hypotenuse
cos θ = base/perpendicular
sin θ = ED/AD or cos θ = AE/AD
AD = ED/sin θ or AD = AE/cos θ
AD = 12/sin θ or AD = 9/cos θ
Hence,
AD = 12/ sin θ or AD = 9/cos θ.
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