a. A boy stands at the point M on the same horizontal level as the foot, T of a vertical building. He observes an object on the top, P of the building at an angle of elevation of 66°. He moves directly backward to a new point C and observes the same object at an angle of 53°. if | MT | = 50 m:
Illustrate the information in a diagram;
bi. Calculate and correct to one decimal place: the height of the building;
bii. Calculate and correct to one decimal place: LINE MC.
a. solution above
bi. From ∆PMT
the height of the building is PT
\(Tan\theta = \frac{opposite}{adjacent}\)
tan66° = \(\frac{PT}{50}\)
Therefore, PT = \(tan 66°\times50 = 112.3m\) ( to 1 d.p)
bii. From ∆PCT
\(Tan53° = \frac{PT}{CT}\)
\(Tan53° = \frac{112.3}{CT}\)
\(Tan53°\times CT = 112.3\)
\(CT = \frac{112.3}{Tan53°}\) = 84. 62m.
MC = CT - MT = 84.62 - 50 = 34.6m
Therefore, MC = 34.6m ( to 1 d.p ).
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