Question

Scientist A standing in North and B in East in terms of position of a rocket of height 75 m that is observed by them with an angle of elevation to the top of the rocket is 45° .What is the distance between them?

answer. - 75√2 m

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Answer 1

Explanation:

FROM SCIENTIST B

$\tan(45) = \frac{75}{disance \: between \: center \: and \: b} \\ 75 = distance \: between \: center \: and \: b$

IF YOU WORK OUT THE SAME THING FOR SCIENTIST A YOU CAN GET THE SAME THING.

THEREFORE DISTANCE BETWEEN CENTRE AND A , DISTANCE BETWEEN CENTRE AND B = 75

THEREFORE IT FORMS A RIGHT ANGLED TRIANGLE

USING PYTHAGORAS THEOREM

DISTANCE BETWEEN A AND B =

$\sqrt{ {75}^{2} + {75}^{2} } = \: 75 \sqrt{2} m$

Answer 2

♣ $\huge{\boxed{\tt \pink{Answer}}}$

FROM SCIENTIST B

$\tan(45) = \frac{75}{disance \: between \: center \: and \: b} \\ 75 = distance \: between \: center \: and \: b$

IF YOU WORK OUT THE SAME THING FOR SCIENTIST A YOU CAN GET THE SAME THING.

THEREFORE DISTANCE BETWEEN CENTRE AND A , DISTANCE BETWEEN CENTRE AND B = 75

THEREFORE IT FORMS A RIGHT ANGLED TRIANGLE

USING PYTHAGORAS THEOREM

DISTANCE BETWEEN A AND B =

$\sqrt{ {75}^{2} + {75}^{2} } = \: 75 \sqrt{2} m$

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