Advertisements
Advertisements
प्रश्न
In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?
उत्तर
cos C = `12/13` .....(i) [Given]
In ∆ABC,
Let ∠ABC = 90°
∴ cos C = `"BC"/"AC"` .....(ii) [By definition]
∴ `"BC"/"AC" = 12/13` ......[From (i) and (ii)]
∴ `24/"AC" = 12/13`
∴ `(24 xx 13)/12` = AC
∴ `312/12` = AC
∴ AC = 26 units
संबंधित प्रश्न
If `cosθ=1/sqrt(2)`, where θ is an acute angle, then find the value of sinθ.
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Without using trigonometric tables evaluate:
`(sin 65^@)/(cos 25^@) + (cos 32^@)/(sin 58^@) - sin 28^2. sec 62^@ + cosec^2 30^@`
if `sin theta = 1/sqrt2` find all other trigonometric ratios of angle θ.
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A + cos A – 1 = 0
Write the maximum and minimum values of sin θ.
If \[\cos \theta = \frac{2}{3}\] find the value of \[\frac{\sec \theta - 1}{\sec \theta + 1}\]
If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
The value of
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
In the following figure the value of cos ϕ is
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
The value of tan 1° tan 2° tan 3°…. tan 89° is
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
