Advertisements
Advertisements
प्रश्न
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Advertisements
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
Use tables to find the acute angle θ, if the value of cos θ is 0.9574
If \[\tan \theta = \frac{4}{5}\] find the value of \[\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\]
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
A, B and C are interior angles of a triangle ABC. Show that
sin `(("B"+"C")/2) = cos "A"/2`
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
`tan 47^circ/cot 43^circ` = 1
