Advertisements
Advertisements
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
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.
APPEARS IN
RELATED QUESTIONS
Calculate the value of A, if (sin A - 1) (2 cos A - 1) = 0
Solve for x : cos `(x/(2)+10°) = (sqrt3)/(2)`
If θ = 30°, verify that: sin2θ = `(2tanθ)/(1 ++ tan^2θ)`
If A = B = 60°, verify that: sin(A - B) = sinA cosB - cosA sinB
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
Evaluate the following: `(tan12°)/(cot78°)`
Evaluate the following: tan(78° + θ) + cosec(42° + θ) - cot(12° - θ) - sec(48° - θ)
If P, Q and R are the interior angles of ΔPQR, prove that `cot(("Q" + "R")/2) = tan "P"/(2)`
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
If A + B = 90°, prove that `(tan"A" tan"B" + tan"A" cot"B")/(sin"A" sec"B") - (sin^2"B")/(cos^2"A")` = tan2A