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A child spots crackers in a cabinet at an angle of elevation of 50. The crackers are 3.5 ft above the child. To the nearest tenth of a foot, what is the horizontal distance between the child and the crackers?

User Jady
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1 Answer

4 votes

the horizontal distance between the child and the crackers is 2.9 ft .

Explanation:

Here we have , A child spots crackers in a cabinet at an angle of elevation of 50. The crackers are 3.5 ft above the child. We need to find what is the horizontal distance between the child and the crackers . Let's find out:

According to question , we have a right angle triangle with parameters as :


Perpendicular =3.5\\Base = ?\\x = 50 , Where x is angle of elevation !

We know that ,
Tanx = (Perpendicular)/(Base)


Tanx = (Perpendicular)/(Base)


Tan50 = (3.5)/(Base)


Base = (3.5)/(Tan50)


Base = (3.5)/(1.19)


Base =2.9ft

Therefore , the horizontal distance between the child and the crackers is 2.9 ft .

User OriEng
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