Advertisements
Advertisements
प्रश्न
In the below given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long. Find the sun’s altitude.
उत्तर
LetAB be the tower and BC be its shadow.
AB = 20, BC = 20√3
In ΔABC,
`tan theta= `
`tan theta=20/(20sqrt3)`
`tan theta=1/sqrt3`
`but,tan30=1/sqrt3`
`theta=30^@`
The Sun is at an altitude of 30º .
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables, evaluate
`sin^2 34^@ + sin^2 56^@ + 2tan 18^@ tan 72^@ - cot^2 30^@`
Without using tables evaluate: 3cos 80°. cosec 10° + 2sin 59° sec 31°
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Without using trigonometric tables, evaluate :
`sin 16^circ/cos 74^circ`
Without using trigonometric tables, prove that:
cos 81° − sin 9° = 0
Without using trigonometric tables, prove that:
cos275° + cos215° = 1
Without using trigonometric tables, prove that:
cos257° − sin233° = 0
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
Prove that:
`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`
Prove that `(sin "A" - cos "A" + 1)/(sin "A" + cos "A" - 1) = 1/(sec "A" - tan "A")`
Given that `tan (θ_1 + θ_2) = (tan θ_1 + tan θ_2)/(1 - tan θ_1 tan θ_2)` Find (θ1 + θ2) when tan θ1 = `1/2 and tan θ_2 = 1/3`.
Solve the following equation: `(cos^2θ - 3 cosθ + 2)/sin^2θ` = 1.
Using trigonometric table evaluate the following:
tan 78°55' - tan 55°18'
`(sin 20°50' + tan 67°40')/(cos 32°20' - sin 15°10')`
The length of a shadow of a tower standing on a level plane is found to be 2y meters longer when the seen's altitude is 30° than when it was 45° prove that the height of the tower is y ( √3 + 1 ) meter.
Given that sin θ = `a/b` then cos θ is equal to ______.
If sin θ = 1, then the value of `1/2 sin(theta/2)`is ______.
