Advertisements
Advertisements
प्रश्न
If sin 3θ = cos (θ – 6°) where 3θ and θ − 6° are acute angles, find the value of θ.
उत्तर
3θ, θ – 6 are an acute angle
We know that sin (90 – θ) = cos θ
sin 3θ = sin (90 – (θ - 6°))
sin 3θ = sin(90 – θ + 6°)
sin 3θ = sin (96° - θ)
3θ = 96° – θ
4θ = 96°
`θ = 96^@/4`
`θ = 24^@`
APPEARS IN
संबंधित प्रश्न
If A = B = 60°, verify that sin (A − B) = sin A cos B − cos A sin B
If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying Sin (20 + 45°) = cos (30 - θ°)
If sin ∝ = `1/2` prove that (3cos∝ - `4cos^2` ∝)=0
If a right ΔABC , right-angled at B, if tan A=1 then verify that 2sin A . cos A = 1
Evaluate:
cos450 cos300 + sin450 sin300
Show that:
(i)` (1-sin 60^0)/(cos 60^0)=(tan60^0-1)/(tan60^0+1)`
In the adjoining figure, ΔABC is right-angled at B and ∠A = 300. If BC = 6cm, find (i) AB, (ii) AC.
If 5 cot θ = 12, find the value of : Cosec θ+ sec θ
If cosB = `(1)/(3)` and ∠C = 90°, find sin A, and B and cot A.
Assertion (A): For 0 < 0 ≤ 90°, cosec θ – cot θ and cosec θ + cot θ are reciprocal of each other.
Reason (R): cot2 θ – cosec2 θ = 1.
