Advertisements
Advertisements
प्रश्न
From the given figure,
find:
(i) cos x°
(ii) x°
(iii) `(1)/(tan^2 xx°) – (1)/(sin^2xx°)`
(iv) Use tan xo, to find the value of y.
Advertisements
उत्तर
(i) cos x° = `(10)/(20)`
cos x° = `(1)/(2)`
(ii) cos x° = `(1)/(2)`
cos x° = cos 60°
x° = 60°
(iii) `(1)/(tan^2x°) – (1)/(sin^2x°) = (1)/(tan^2 60°) – (1)/(sin^2 60°)`
= `(1)/(sqrt3)^2 – (1)/(sqrt3/2)^2`
= `(1)/(3) – (4)/(3)`
= – 1
(iv) tan x° = tan 60°
= `sqrt3`
We know that tan x° = `"AB"/"BC"`
⇒ tan x° = `"y"/(10)`
⇒ y = 10 tan x°
⇒ y = 10 tan 60°
⇒ y = 10`sqrt3`
APPEARS IN
संबंधित प्रश्न
Calculate the value of A, if (sin A - 1) (2 cos A - 1) = 0
Solve for x : cos `(x/(2)+10°) = (sqrt3)/(2)`
Solve for x : sin2 x + sin2 30° = 1
Find the value of 'A', if cot 3A = 1
If sin α + cosβ = 1 and α= 90°, find the value of 'β'.
Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`
Evaluate the following: `(sec34°)/("cosec"56°)`
Evaluate the following: sec16° tan28° - cot62° cosec74°
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cosec64° + sec70°
If secθ= cosec30° and θ is an acute angle, find the value of 4 sin2θ - 2 cos2θ.
