a tower is 100 m high . find the angpe of elevation of its top from a point 100√3 away from its foot
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Given that:
Height of a tower = 100m
Distance from foot of tower to angle of elevation point = 100root3 m
tan theta = opp side/ adj side
tan theta = 100/100 root3
tan theta = 1/root 3
we onow that tan60 = 1/ root3
therefore theta =60°
Height of a tower = 100m
Distance from foot of tower to angle of elevation point = 100root3 m
tan theta = opp side/ adj side
tan theta = 100/100 root3
tan theta = 1/root 3
we onow that tan60 = 1/ root3
therefore theta =60°
vaibhav4961:
no bro the answer is 30 but how i dont know
i had also done it
yeah yeah 1/√3 is 30 this was the mistake
ohoo sorry
yeah sorry answer was root 3.. for that it was 60°
Answered by
2
tan. thit = p/base
so,100/100root3
now the value of root 3=60
so,100/100root3
now the value of root 3=60
bro 1/√3=30
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