Advertisements
Advertisements
प्रश्न
Find the value 'x', if:
Advertisements
उत्तर
In right ΔABC,
sin45° = `"BC"/"AC"`
⇒ `(1)/sqrt(2) = (15)/x`
⇒ x = `15sqrt(2)"cm"`.
APPEARS IN
संबंधित प्रश्न
Solve for x : sin2 60° + cos2 (3x- 9°) = 1
Find the value of 'A', if cosec 3A = `(2)/sqrt(3)`
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
The perimeter of a rhombus is 100 cm and obtuse angle of it is 120°. Find the lengths of its diagonals.
Evaluate the following: `(2sin28°)/(cos62°) + (3cot49°)/(tan41°)`
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: sin65° + cot59°
If sin(θ - 15°) = cos(θ - 25°), find the value of θ if (θ-15°) and (θ - 25°) are acute angles.
If cosθ = sin60° and θ is an acute angle find the value of 1- 2 sin2θ
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
