-->

Circles

Question
CBSEENMA10008320

From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 30° and 45° respectively. Find the height of the hill.

Solution

Let CD be the hill of height h km. Let A and B be two stones due east of the hill at a distance of 1 km. from each other. It is also given that the angles of depression of. stones A and B from the top of a hill be 30° and 45° respectively.
Let    BC = x km
In right triangle BCD, we have
space space space space tan space 45 degree space space equals space CD over BC
rightwards double arrow space space space space space 1 space equals space straight h over straight x
rightwards double arrow space space space space space space straight x space equals space straight h space space space space space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
In right triangle ACD, we have
tan space 30 degree space equals space CD over AC
rightwards double arrow space space space fraction numerator 1 over denominator square root of 3 end fraction space equals space fraction numerator straight h over denominator 1 plus straight x end fraction
rightwards double arrow space space 1 plus straight x space equals space square root of 3 straight h
rightwards double arrow space space space straight x space equals space square root of 3 straight h space minus space 1 space space equals space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
Comparing (i) and (ii), we get
straight h space equals space square root of 3 space straight h space minus space 1
rightwards double arrow space square root of 3 straight h end root minus straight h space equals space 1
rightwards double arrow space straight h open parentheses square root of 3 minus 1 close parentheses space equals space 1
rightwards double arrow space straight h space equals space fraction numerator 1 over denominator square root of 3 minus 1 end fraction straight x fraction numerator square root of 3 plus 1 over denominator square root of 3 plus 1 end fraction
rightwards double arrow space space straight h space equals space fraction numerator square root of 3 plus 1 over denominator left parenthesis square root of 3 right parenthesis squared minus left parenthesis 1 right parenthesis squared end fraction
rightwards double arrow space space space space equals space space fraction numerator square root of 3 plus 1 over denominator 3 minus 1 end fraction
rightwards double arrow space space space fraction numerator square root of 3 plus 1 over denominator 2 end fraction equals fraction numerator 1.732 plus 1 over denominator 2 end fraction equals fraction numerator 2.732 over denominator 2 end fraction
equals space space 1.365 space km.
Hence, the height of hill is 1.365.

Some More Questions From Circles Chapter