Advertisements
Advertisements
प्रश्न
Find the magnitude of angle A, if 2 cos2 A - 3 cos A + 1 = 0
Advertisements
उत्तर
2 cos2 A – 3 cos A + 1 = 0
2 cos2 A – cos A - 2cosA +1 = 0
cos A(2cos A – 1) – (2 cos A – 1) = 0
( 2 cosA – 1) (cos A – 1) = 0
2 cos A – 1 = 0 aand cos A – 1 = 0
cos A = `(1)/(2)` and cos A = 1
A = 60° and A = 0°
APPEARS IN
संबंधित प्रश्न
In ΔABC, ∠B = 90° , AB = y units, BC = `(sqrt3)` units, AC = 2 units and angle A = x°, find:
- sin x°
- x°
- tan x°
- use cos x° to find the value of y.
Solve the following equation for A, if 2 sin 3 A = 1
Calculate the value of A, if (cosec 2A - 2) (cot 3A - 1) = 0
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
Solve for x : cos2 30° + cos2 x = 1
If θ < 90°, find the value of: sin2θ + cos2θ
Find lengths of diagonals AC and BD. Given AB = 24 cm and ∠BAD = 60°.
Evaluate the following: `(3sin37°)/(cos53°) - (5"cosec"39°)/(sec51°) + (4tan23° tan37° tan67° tan53°)/(cos17° cos67° "cosec"73° "cosec"23°)`
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
