Question 33

Let QRS be a triangular park in xy-plane with side RS = 375 meters and angle QRS = 90°. A pole PQ vertical to the xy-plane is fixed at Q with height PQ = h. if tan PRQ = $$\frac{17}{25}$$ and tan PSQ = $$\frac{8}{25}$$ then the value of h (in meters) is

Solution

$$\frac{h}{SQ}=\frac{8}{25}$$

$$\frac{h}{RQ}=\frac{17}{25}$$

=> $$SQ^2=\frac{25^2}{8^2}h^2$$

$$RQ^2=\frac{25^2}{17^2}h^2$$

$$SQ^2=RQ^2+375^2$$

Upon solving, h = 136

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