A hot air balloon is spotted simultaneously by a man and a woman 3 ½ miles apart. If the angle of elevation from the man to the
balloon is 27° and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon in feet.
1 answer:
See the picture attached to better understand the problem
we know that
in the right triangle ABD
tan 27°=BD/AD-----> BD=(3.5-x)*tan 27°------> equation 1
in the right triangle BDC
tan 41°=BD/DC-----> BD=x*tan 41°-----> equation 2
equate equation 1 and equation 2
(3.5-x)*tan 27°=x*tan 41°----> 3.5*tan 27-x*tan 27=x*tan 41
x*[tan 27+tan 41]=3.5*tan 27
x=3.5*tan 27/[tan 27+tan 41]-----> x=1.29 miles
BD=x*tan 41-----> BD=1.29*tan 41°----> BD=1.124 miles
convert miles to ft
1 mile-------> 5280 ft
1.124 miles----> x
x=1.124*5280-----> x=5934.72 ft
the answer is
<span>the altitude of the balloon in feet is 5934.72</span>
we know that
in the right triangle ABD
tan 27°=BD/AD-----> BD=(3.5-x)*tan 27°------> equation 1
in the right triangle BDC
tan 41°=BD/DC-----> BD=x*tan 41°-----> equation 2
equate equation 1 and equation 2
(3.5-x)*tan 27°=x*tan 41°----> 3.5*tan 27-x*tan 27=x*tan 41
x*[tan 27+tan 41]=3.5*tan 27
x=3.5*tan 27/[tan 27+tan 41]-----> x=1.29 miles
BD=x*tan 41-----> BD=1.29*tan 41°----> BD=1.124 miles
convert miles to ft
1 mile-------> 5280 ft
1.124 miles----> x
x=1.124*5280-----> x=5934.72 ft
the answer is
<span>the altitude of the balloon in feet is 5934.72</span>
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