Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर

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
संबंधित प्रश्न
Solve the following equation for A, if sec 2A = 2
Evaluate the following: `((sin3θ - 2sin4θ))/((cos3θ - 2cos4θ))` when 2θ = 30°
Evaluate the following: `((1 - cosθ)(1 + cosθ))/((1 - sinθ)(1 + sinθ)` if θ = 30°
Find the value 'x', if:
Find the value of 'y' if `sqrt(3)` = 1.723.
Given your answer correct to 2 decimal places.
Evaluate the following: `(sec32° cot26°)/(tan64° "cosec"58°)`
Evaluate the following: `(3sin37°)/(cos53°) - (5"cosec"39°)/(sec51°) + (4tan23° tan37° tan67° tan53°)/(cos17° cos67° "cosec"73° "cosec"23°)`
Evaluate the following: `(sin0° sin35° sin55° sin75°)/(cos22° cos64° cos58° cos90°)`
If cos3θ = sin(θ - 34°), find the value of θ if 3θ is an acute angle.
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
