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
Solve.
sin42° sin48° - cos42° cos48°
Evaluate.
`cos^2 26^@+cos65^@sin26^@+tan36^@/cot54^@`
Evaluate:
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Use tables to find the acute angle θ, if the value of cos θ is 0.9574
Use tables to find the acute angle θ, if the value of cos θ is 0.6885
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
The value of tan 72° tan 18° is
