Question

The angles of elevation of the top of the tower from two points at a distance of 150 and 100 m
​

Answer 1

AnSwer:

⠀⠀⠀⠀

⋆ Reference of image is shown in diagram:

⠀⠀⠀⠀

$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\qbezier(12,1)(11,1.4)(8,2.9)\put(7.5,2){\sf{\large{h m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\qbezier(11,1)(10.8,1.2)(11.1,1.4)\put(9.6,1.2){\sf\large{\theta }}\put(10.2,1.3){\sf\footnotesize{(90^ \circ - \theta) }}\put(11.9,.7){\sf\large A}\put(7.9,3){\sf\large D}\put(10.4,.7){\sf\large B}\put(7.9,.7){\sf\large C}\put(8.1,0.8){\vector(3,0){0.8}}\put(10.3,0.8){\vector( - 3,0){0.8}}\put(8.95,0.75){\sf 100 m}\put(8.1,0.4){\vector(3,0){1.6}}\put(12,0.4){\vector( - 3,0){1.6}}\put(9.8,0.35){\sf 150 m}\end{picture}$

⠀⠀⠀⠀

• Let height of the tower, CD = h m.
• $\sf Let\:\angle\:CBD = \theta\:then\: \angle\:CAD = (90 - \theta)$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

$\bigstar\:{\underline{\sf{In\: right\:angled\:\triangle\:ACD,}}}$

$:\implies\sf tan(90 - \theta) = \dfrac{CD}{AC}\\ \\ \\ :\implies\sf tan(90 - \theta) = \dfrac{h}{150}\\ \\ \\ :\implies\sf cot \theta = \dfrac{h}{150}\qquad\qquad\bigg\lgroup\bf eq\:(1) \bigg\rgroup$

$\bigstar\:{\underline{\sf{In\:right\:angled \:\triangle\:ACD,}}}$

$:\implies\sf tan \theta = \dfrac{CD}{BC}\\ \\ \\ :\implies\sf tan \theta= \dfrac{h}{100}\qquad\qquad\bigg\lgroup\bf eq\:(2) \bigg\rgroup$

$\bigstar\:{\underline{\sf{Now,\: Multiplying\:eq\;(1)\:and\;eq\;(2),}}}$

$:\implies\sf cot \theta \times tan \theta = \dfrac{h}{150} \times \dfrac{h}{100}\\ \\ \\ :\implies\sf \cancel{cot \theta} \times \dfrac{1}{ \cancel{cot \theta}} = \dfrac{h}{150} \times \dfrac{h}{100}\\ \\ \\ :\implies\sf 1 = \dfrac{h^2}{15000}\\ \\ \\ :\implies\sf h^2 = 15000\\ \\ \\ :\implies\sf \sqrt{h^2} = \sqrt{15000}\\ \\ \\ :\implies\sf h = \sqrt{6 \times 25 \times 100}\\ \\ \\ :\implies\sf h = \sqrt{6 \times (5)^2 \times (10)^2}\\ \\ \\ :\implies\sf h = 5 \times 10 \times \sqrt{6}\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{h = 50 \sqrt{6}\;m}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Height\;of\;the\;tower\;is\; \bf{50 \sqrt{6}}.}}}$

$\qquad\qquad\qquad\qquad\dag\;{\underline{\underline{\bf{\pink{Hence\:Proved!}}}}}$