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
APPEARS IN
संबंधित प्रश्न
Solve.
`cos55/sin35+cot35/tan55`
Evaluate:
cosec (65° + A) – sec (25° – A)
Find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° – 3A) . cosec 42° = 1
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
Evaluate: `(sin 80°)/(cos 10°)`+ sin 59° sec 31°
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.
