Question

two men of the same side of the tower and in the same straight line with its base notice the angle of elevation of the top of the tower to be 30° and 60°.if the height of the tower is 150m. find the distance between two men

Answer 1

Answer .

The two persons are standing on point D and C respectively.

Let the tower as AB and the angle ACB is 60 ° and ADB is 30° . Here Angle B is 90° .BC is Y and DB is x.

In Right angled∆ ABC

<B = 90° .

tan 60° = AB / CB.

√3 = 150/ y

y. = 150 /√3. ( i)

In Right angled ∆ ADB .

<B = 90°

tan 30° = AB/DB

1/√3. = 150/x

x. = 150√3.

We have to find distance between the two persons that is DC .

DC = x - y

DC.= 150√3 - 150/√3

$dc = \frac{150(3) - 150}{ \sqrt{3} }$

$dc = \frac{150(3 - 1)}{ \sqrt{3} }$

DC = 150(2) /√3

DC = 300/√3

rationalize .

DC = 300× √3/√3×√3

DC = 100√3 m.

So distance between the two persons is 100 √3 m .

Answer 2

Answer:

91.38

Step-by-step explanation:

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Two person standing on the same side of a tower in a straight line with it measure the angle of elevation of the top of the tower as 25˚ and 50˚ respectively.

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asked Apr 1, 2019 in Class X Maths by priya12 (-12,631 points)

Two person standing on the same side of a  tower in a straight line with it measure the angle of elevation of the top of the tower as 25˚ and 50˚ respectively. If the height of the tower is 70 m find the distance between the two person.

trigonometry

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1 Answer

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answered Apr 1, 2019 by navnit40 (-4,939 points)

Let CD be the distance between the two persons

In  △ABC,

cot 50˚ = BC/AB

cot (90˚ – 40˚) = BC/70

tan 40˚ = BC/70

BC =  70 tan 40˚

= 70 × 0.8391 = 58.74 m

In △ABD,

 cot 25˚  = BD/AB

 cot (90˚ – 65˚) = BD/70

tan 65˚ = BD/70

BD = 70 tan 65˚

= 70 × 2.11451

= 150.12 m

CD = 150.12 – 58.74

= 91.38 m