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Question
In the given figure, PQ = 6 cm, RQ = x cm and RP = 10 cm, find
a. cosθ
b. sin2θ- cos2θ
c. Use tanθ to find the value of RQ
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Solution

a cosθ = `"Base"/"Hypotenuse"`
⇒ cosθ
= `"PQ"/"PR"`
= `(6)/(10)`
= `(3)/(5)`
b. sin2θ + cos2θ = 1
⇒ `sin^2θ + (3/5)^2` = 1
⇒ `sin^2θ + (9)/(25)` = 1
⇒ sin2θ = `1 - (9)/(25) = (16)/(25)`
⇒ sinθ = `(4)/(5)`
∴ sin2θ - cos2θ
= `(6)/(25) - (9)/(5)`
= `(7)/(25)`
c. tanθ
= `(sinθ)/(cosθ)`
= `(4/5)/(3/5)`
= `(4)/(3)`
But,
tanθ = `"Perpendicular"/"Base" = "RQ"/"PQ"`
⇒ `"RQ"/"PQ" = (4)/(3)`
⇒ `"RQ"/(6) = (4)/(3)`
⇒ RQ
= `(4 xx 6)/(3)`
= 8cm.
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