Question

The angle of elevation of top point P of vertical tower PQ of height h from point A is 45 and from point B is at a distance d from point A measure along the line AB which makes an angle 30 with AQ. Prove that d=h( Ö 3-1)

Answer 1

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Answer 2

ANSWER: $\quad \left( \sqrt { 3 } \quad -\quad 1 \right) h$

PQ = h

AB = d

            $\tan { \theta } \quad =\quad \frac { opposite }{ adjacent }$

            $\tan { 45 } \quad =\quad \frac { PQ }{ AQ }$

            $1\quad =\quad \frac { h }{ AQ }$

            $AQ\quad =\quad h$

            $\tan { 30 } \quad =\quad \frac { PQ }{ BQ }$

            $=\quad \frac { h }{ BA\quad +\quad AQ }$

            $\frac { 1 }{ \sqrt { 3 }  } \quad =\quad \frac { h }{ d\quad +\quad h }$

            $d\quad +\quad h\quad =\quad \sqrt { 3 } h\quad$

            $d\quad =\quad \sqrt { 3 } h\quad -\quad h$

            $d\quad =\quad \left( \sqrt { 3 } \quad -\quad 1 \right) h$

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