Question

A fire in a building B is reported on teleported on telephone to two fire stations P and Q, 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60° to the road and Q observes that it is at an angle of 45° to the road. Which station should send its team and how much will this team have to travel?

Answer 1

Fire station P will send their team , and team will travel 7.33 km of distance

Step-by-step explanation:

Given as :

Two fire stations P and Q, 20 km apart from each other on a straight road.

The distance between point P and point Q = 20 km

The angle of elevation of point P with ground  = 60°

The angle of elevation of point Q with ground  = 45°

Let The distance cover by fire station Q = x  km

Let The distance cover by fire station P = ( 20 - x)  km

The height of angle of elevation of both stations = h km

According to question

Tan angle = $\dfrac{perpendicular}{base}$

i.e Tan 45° = $\dfrac{QY}{QO}$

Or, 1 = $\dfrac{h}{x}$

i.e  h = x               ........1

Again

Tan angle = $\dfrac{perpendicular}{base}$

i.e Tan 60° = $\dfrac{PX}{PO}$

Or, √3 = $\dfrac{h}{20-x}$

i.e  h = ( 20 - x) √3           ....2

From eq 1 and 2

Put the value of h in eq 2

x = ( 20 - x) √3

Or, x + x√3 = 20√3

Or, x = $\dfrac{20\sqrt{3} }{1+\sqrt{3} }$

Or,  x = $\dfrac{(20\sqrt{3})(\sqrt{3} -1) }{2}$

Or, x = 30 - 10√3

∴    x = 12.67  km

So, The distance that fire station Q travel = x = 12.67 km

So, The distance that fire station P travel = 20 - x = 20 - 12.67 = 7.33 km

Hence , fire station P will send their team , and team will travel 7.33 km of distance . Answer

     

Answer 2

Answer:

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