A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of $30°$ and $60°$, which are approaching the foot of the to

Question

A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of $30°$ and $60°$, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use $\sqrt{3} = 1.73$)

Accepted Answer

Step 1: Tower AB = 75 m with two cars at P and Q, angles of depression 60 degrees and 30 degrees
Step 2: AB = Height of tower = 75 m. P, Q are positions of cars. $\angle XBQ = \angle BQA = 30°$, $\angle XBP = \angle BPA = 60°$
Step 3: In $	riangle APB$: $	an 60° = \frac{75}{AP} \Rightarrow AP = \frac{75}{\sqrt{3}} = 25\sqrt{3}$
Step 4: In $	riangle AQB$: $	an 30° = \frac{75}{AQ} \Rightarrow AQ = 75\sqrt{3}$
Step 5: Distance between the cars $= PQ = AQ - AP = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} = 50 	imes 1.73 = 86.5$ m

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