Question

The length of the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45 to 30. Then the height of the tower is

Answer 1

DIAGRAM:

⠀

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\qbezier(12,1)(11,1.4)(8,2.9)\put(7.7,2){\sf{\large{h}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\qbezier(11,1)(10.8,1.2)(11.1,1.4)\put(9.4,1.2){\sf\large{45^{\circ} }}\put(10.5,1.2){\sf\large{30^{\circ} }}\put(11.9,.7){\sf\large D}\put(7.9,3){\sf\large B}\put(10.4,.7){\sf \large C}\put(7.9,.7){\sf\large A}\put(8.2,0.8){\vector(1,0){0.7}}\put(9,.7){\sf\large x m}\put(10.2,0.8){\vector( - 1,0){0.7}}\put(10.6,0.8){\vector(1,0){0.2}}\put(10.9,.7){\sf\large 10 m}\put(11.8,0.8){\vector( - 1,0){0.2}}\end{picture}$

⠀

Let AB be the tower and AC and AD be it's shadows when the angles of elevation of the sun are 40° and 30° respectively. Then, CD = 10 metres.

Let h be the height of tower and let AC be x metres.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

${\underline{\sf{\bigstar\;In\; \triangle\;CAB,\;}}}$

$\qquad\sf tan\;45^\circ = \dfrac{AB}{AC}$

$\qquad:\implies\sf 1 = \dfrac{h}{x}$

$\: \qquad:\implies\sf x = h\qquad\qquad\bigg\lgroup\bf eq\;(1) \bigg\rgroup$

${\underline{\sf{\bigstar\;Now,\;In\; \triangle\;CAB,\;}}}$

$\quad\qquad\sf tan\;30^\circ = \dfrac{AB}{AD}$

$\quad:\implies\sf \dfrac{1}{ \sqrt{3}} = \dfrac{h}{x + 10}$

$\: \quad:\implies\sf x + 10 = \sqrt{3} h\qquad\qquad\bigg\lgroup\bf eq\;(2) \bigg\rgroup$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

✇ Now, Substituting the value of x obtained from eq (1) and eq (2),

$\quad\qquad\qquad\sf h + 10 = \sqrt{3}$

$\qquad:\implies\sf h( \sqrt{3} - 1) = 10$

$\qquad\quad:\implies\sf h = \dfrac{10}{ \sqrt{3} - 1}$

$:\implies\sf h = \dfrac{10}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1}{ \sqrt{3} + 1}$

$\qquad:\implies\sf h = 10 \bigg\lgroup\sf \dfrac{ \sqrt{3} + 1}{2} \bigg\rgroup$

$\qquad\quad:\implies\sf h = 5( \sqrt{3} + 1)$

$\quad\qquad:\implies\sf h = 5(1.73 + 1)$

$\qquad\qquad:\implies{\underline{\boxed{\frak{\purple{13.65\;m}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;height\;of\;the\;tower\;is\; \bf{13.65\;metres}.}}}$

Answer 2

LINE OF SIGHT: The line of sight is a line drawn from the eye of an observer to the point in the object viewed by the observer.

ANGLE OF ELEVATION: The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal , when it is above the horizontal level.

ANGLE OF DEPRESSION:The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal , when it is below the horizontal level.

•Angle of elevation and depression are always acute angles.

•If the observer moves towards the perpendicular line(Tower/ building) then angle of elevation increases and if the observer move away from the perpendicular line(Tower/ building) angle of elevation decreases.