Advertisements
Advertisements
प्रश्न
Find the value of 'x' in each of the following:
Advertisements
उत्तर
From the figure, we have
sin x = `"BC"/"AC"`
⇒ sin x = `(15/sqrt(2))/(15)`
⇒ sin x = `(1)/sqrt(2)`
⇒ sin x = sin45°
⇒ x = 45°.
APPEARS IN
संबंधित प्रश्न
Find the magnitude of angle A, if 2 sin A cos A - cos A - 2 sin A + 1 = 0
Solve the following equation for A, if 2 sin A = 1
Solve for x : sin (x + 10°) = `(1)/(2)`
Solve for x : sin2 x + sin2 30° = 1
Find the value of 'A', if `sqrt(3)cot"A"` = 1
In right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if:
a. AB is `sqrt(3)` times of BC.
B. BC is `sqrt(3)` times of BC.
In the given figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically upwards and then 20 km at 60° to the vertical. PQ represents the first stage of the journey and QR the second. S is a point vertically below R on the horizontal level as P, find:
a. the height of the rocket when it is at point R.
b. the horizontal distance of point S from P.
Evaluate the following: cot27° - tan63°
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cos84° + cosec69° - cot68°
If A + B = 90°, prove that `(tan"A" tan"B" + tan"A" cot"B")/(sin"A" sec"B") - (sin^2"B")/(cos^2"A")` = tan2A
