Advertisements
Advertisements
प्रश्न
If cos A = `13/14` and cos B = `1/7` where A, B are acute angles prove that A – B = `pi/3`
उत्तर
cos A = `13/14`, cos B = `1/7`
sin A = `sqrt(1 - cos^2"A")`
`= sqrt(1 - 1/49) = sqrt(48/49) = sqrt48/7 = (4sqrt3)/7`
cos(A – B) = cos A cos B + sin A sin B
`= 13/14 xx 1/7 + (3sqrt3)/14 xx (4sqrt3)/7`
`= 13/98 + 36/98 = 49/98 = 1/2`
cos(A – B) = cos 60°
A – B = 60° = `pi/3`
APPEARS IN
संबंधित प्रश्न
Find the value of the following:
sin 76° cos 16° – cos 76° sin 16°
Find the value of the following:
`sin pi/4 cos pi/12 + cos pi/4 sin pi/12`
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22`1/2`.
Prove that `(sin ("B - C"))/(cos "B" cos "C") + (sin ("C - A"))/(cos "C" cos "A") + (sin ("A - B"))/(cos "A" cos "B")` = 0
Find the value of tan `pi/8`.
If tan α = `1/7`, sin β = `1/sqrt10`. Prove that α + 2β = `pi/4` where 0 < α < `pi/2` and 0 < β < `pi/2`.
Prove that `2 sin^2 (3pi)/4 + 2 cos^2 pi/4 + 2 sec^2 pi/3` = 10
If sin A = `1/3`, sin B = `1/4` then find the value of sin (A + B) where A and B are acute angles.
If cos (α + β) = `4/5` and sin (α - β) = `5/13` where (α + β) and (α - β) are acute, then find tan 2α.
The value of `(2 tan 30^circ)/(1 + tan^2 30^circ)` is: