Advertisements
Advertisements
प्रश्न
If Sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A =?
उत्तर
Cos 𝜃 = sin (90° - 𝜃)
⇒ Cos (A – 26) = sin (90° −(A− 26°))
⇒ Sin 3A = sin (90° − (A – 26))
Equating angles on both sides
3A = 90° − A + 26°
`4A = 116° A = 116/4 = 29^@`
`∴ A = 29^@`
APPEARS IN
संबंधित प्रश्न
Evaluate the following in the simplest form: sin 60º cos 45º + cos 60º sin 45º
Evaluate the following:
2tan2 45° + cos2 30° − sin2 60°
Evaluate: `(3 cos 55^@)/(7 sin 35^@) - (4(cos 70 cosec 20^@))/(7(tan 5^@ tan 25^@ tan 45^@ tan 65^@ tan 85^@))`
For any angle θ, state the value of: sin2 θ + cos2 θ
Given A = 60° and B = 30°,
prove that : cos (A + B) = cos A cos B - sin A sin B
If A = 30o, then prove that :
2 cos2 A - 1 = 1 - 2 sin2A
Without using tables, find the value of the following: `(4)/(cot^2 30°) + (1)/(sin^2 60°) - cos^2 45°`
Prove that : sec245° - tan245° = 1
If A = 30° and B = 60°, verify that: cos (A + B) = cos A cos B - sin A sin B
Verify cos3A = 4cos3A – 3cosA, when A = 30°
