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
Calculate the value of A, if (sin A - 1) (2 cos A - 1) = 0
Solve the following equations for A, if `sqrt3` tan A = 1
Find the magnitude of angle A, if 2 cos2 A - 3 cos A + 1 = 0
Evaluate the following: `((1 - cosθ)(1 + cosθ))/((1 - sinθ)(1 + sinθ)` if θ = 30°
Find the value 'x', if:
In the given figure, if tan θ = `(5)/(13), tan α = (3)/(5)` and RS = 12m, find the value of 'h'.
Evaluate the following: `(cos34° cos35°)/(sin57° sin56°)`
Evaluate the following: `(5cot5° cot15° cot25° cot35° cot45°)/(7tan45° tan55° tan65° tan75° tan85°) + (2"cosec"12° "cosec"24° cos78° cos66°)/(7sin14° sin23° sec76° sec67°)`
If P, Q and R are the interior angles of ΔPQR, prove that `cot(("Q" + "R")/2) = tan "P"/(2)`
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
