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The figure shows a weather balloon at point P. How high is the point P of the weather balloon from a point R on the ground?
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How high is the point P of the weather balloon from a point R on the ground?
@lalaly cud u help
since u are given an angle and a side of a right triangle use \[\tan(\theta)=\frac{opposite side}{adjacent}\]\[\tan(32)=\frac{RP}{5}\]
now solve for RP
5 by sec 32 degrees miles 5 by cot 32 degrees miles 5 cosec 32˚ miles 5 cot 32˚ miles
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@lalaly
Oh never mind i got it lol its the second choice 5 by cot32
lol! thank u
@lalaly
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