Advertisements
Advertisements
प्रश्न
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
उत्तर
cos (90° – A) . sec 77° = 1
`sinA. 1/(cos77^circ) = 1`
sin A = cos 77°
= cos (90° – 13°)
= sin 13°
A = 13°
APPEARS IN
संबंधित प्रश्न
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Find the value of x, if sin 3x = 2 sin 30° cos 30°
Use tables to find sine of 62° 57'
Evaluate:
cos 40° cosec 50° + sin 50° sec 40°
If \[\sec\theta = \frac{13}{12}\], find the values of other trigonometric ratios.
Given
\[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\] what is the value of \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\]
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
If 3 cos θ = 5 sin θ, then the value of
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?
