The top of a tree of height 3 mm^(1/2) units is seen from distances m units and n units...
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The top of a tree of height 3 mm^(1/2) units is seen from distances m units and n units from the base of the tree on the ground (m < n). What is the difference between the angle of elevation from m and n respectively?
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The angles of elevation correspond to tan(theta_1) = sqrt(3) (60 degrees) and tan(theta_2) = 1/sqrt(3) (30 degrees). The difference between the angles is 60 - 30 = 30 degrees.
Step-by-step Derivation:
Step 1: The height of the tree relates to m and n as h = sqrt(3)*m, and n = 3m.
Step 2: Angle from distance m: tan(theta_1) = h / m = sqrt(3) => theta_1 = 60 degrees.
Step 3: Angle from distance n: tan(theta_2) = h / (3m) = sqrt(3) / 3 = 1/sqrt(3) => theta_2 = 30 degrees.
Step 4: Difference = theta_1 - theta_2 = 60 - 30 = 30 degrees.