Advertisements
Advertisements
प्रश्न
Use tables to find sine of 62° 57'
Advertisements
उत्तर
sin 62° 57' = sin (62° 54' + 3')
= 0.8902 + 0.0004
= 0.8906
संबंधित प्रश्न
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
Use tables to find the acute angle θ, if the value of tan θ is 0.4741
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 \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
The value of cos 1° cos 2° cos 3° ..... cos 180° is
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?
