Advertisements
Advertisements
Question
If sin θ =7/25, where θ is an acute angle, find the value of cos θ.
Solution
`sintheta=7/25`
we know `sin theta="Opposite"/"Hypotunes"=p/h`
`therefore p/h=7/25` [∵ Opposite = Perpendicular = p]
p=7k, h=25k
Let the adjacent (base) side be b.
Thus `b=sqrt((25k)^2-(7k)^2)=24k`
`costheta=(24k)/(25k)=24/25`
APPEARS IN
RELATED QUESTIONS
If the angle θ= –60º, find the value of cosθ.
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ
solve.
sec2 18° - cot2 72°
Evaluate.
`(2tan53^@)/(cot37^@)-cot80^@/tan10^@`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Use tables to find sine of 62° 57'
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
If 0° < A < 90°; find A, if `(cos A )/(1 - sin A) + (cos A)/(1 + sin A) = 4`
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If 8 tan x = 15, then sin x − cos x is equal to
If tan2 45° − cos2 30° = x sin 45° cos 45°, then x =
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
If sec A + tan A = x, then sec A = ______.
The value of (tan1° tan2° tan3° ... tan89°) is ______.