From the top of a hill the angles of depression ot two consecutive kilometer stones are found to be 300 and 400 respectively. Find the distance of the two stones from the foot of the hill.
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Let AB is the height of the hill and two stones are C and D respectively where depression is 45 degree and 30 degree. The distance between C and D is 1 km.
Here depression and hill has formed right angle triangles with the base. We have to find the height of the hill with this through trigonometry.
In triangle ABC, tan 45 = height/base = AB/BC
or, 1 = AB/BC [ As tan 45 degree = 1]
or, AB = BC ..........(i)
Again, triangle ABD, tan 30 = AB/BD
or, 1√3=ABBC+CD [tan 30 = 1√3 =1/1.732]
or, 11.732=ABAB+1 [ As AB = BC from (i) above]
or, 1.732 AB = AB +1
or, 1.732 AB - AB = 1
or, AB(1.732-1) = 1
or, AB * 0.732 = 1
or AB = 1/0.732 = 1.366
Hence height of the hill 1.366 km
Here depression and hill has formed right angle triangles with the base. We have to find the height of the hill with this through trigonometry.
In triangle ABC, tan 45 = height/base = AB/BC
or, 1 = AB/BC [ As tan 45 degree = 1]
or, AB = BC ..........(i)
Again, triangle ABD, tan 30 = AB/BD
or, 1√3=ABBC+CD [tan 30 = 1√3 =1/1.732]
or, 11.732=ABAB+1 [ As AB = BC from (i) above]
or, 1.732 AB = AB +1
or, 1.732 AB - AB = 1
or, AB(1.732-1) = 1
or, AB * 0.732 = 1
or AB = 1/0.732 = 1.366
Hence height of the hill 1.366 km
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