Advertisements
Advertisements
प्रश्न
If A and B are acute angles such that sin (A – B) = 0 and 2 cos (A + B) – 1 = 0, then find angles A and B.
उत्तर
sin (A – B) = 0 ...(Given)
`\implies` sin (A – B) = sin 0° ...(∵ sin 0° = 0)
`\implies` A – B = 0 ...(i)
and 2 cos (A + B) – 1 = 0 ...(Given)
`\implies` 2 cos (A + B) = 1
`\implies` cos (A + B) = `1/2`
`\implies` cos (A + B) = cos 60° ...`(∵ cos 60^circ = 1/2)`
`\implies` A + B = 60° ...(ii)
On adding equations (i) and (ii), we get
2A = 60°
`\implies` A = 30°
Put this value in equation (i), we get
30° – B = 0
`\implies` B = 30°
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
2tan2 45° + cos2 30° − sin2 60°
`(2 tan 30°)/(1-tan^2 30°)` = ______.
State whether the following is true or false. Justify your answer.
The value of sinθ increases as θ increases.
If sin x = cos y, then x + y = 45° ; write true of false
find the value of: cos2 60° + sec2 30° + tan2 45°
Given A = 60° and B = 30°,
prove that : cos (A - B) = cos A cos B + sin A sin B
If A = 30°;
show that:
`(1 – cos 2"A")/(sin 2"A") = tan"A"`
Prove that : sec245° - tan245° = 1
Prove that: `((cot30° + 1)/(cot30° -1))^2 = (sec30° + 1)/(sec30° - 1)`
Find the value of x in the following: `sqrt(3)sin x` = cos x
