Advertisements
Advertisements
प्रश्न
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
Advertisements
उत्तर
Let the width of the river be w.
In ΔABC,
tan 60° = `"AB"/"BC"`
⇒ `sqrt3` = `60/w`
⇒ w = `60/(sqrt3) = (60sqrt3)/3= 20sqrt3`
In △AED,
tan30° = `"AE"/"ED"`
⇒ `1/(sqrt3) = "AE"/w`
⇒ `1/(sqrt3) = "AE"/(20sqrt3)`
⇒ AE = 20
Height of pole CD = AB − AE
= 60 − 20 = 40 m.
Thus, width of river is `20sqrt3` = 20 x 1.732 = 34.64 m
Height of pole = 40 m.
संबंधित प्रश्न
Prove the following trigonometric identities.
tan2 θ − sin2 θ = tan2 θ sin2 θ
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
If tan α + cot α = 2, then tan20α + cot20α = ______.
If cosA + cos2A = 1, then sin2A + sin4A = 1.
