Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building​

Answer 1

Answer:

The height of the tower is 118. 3 metres and distance between them is 68.3 metres

Step-by-step explanation:

In this question :

AB = h

AE = h-50

CD = 50 metres

BD = EC = X

In triangle ACE,

$\tan(\theta) = \frac{perpendicular}{base}$

Here,

$\tan(45) = \frac{AE}{ EC}$

Tan 45° = 1

$1 = \frac{h - 50}{x}$

x = h - 50

Now in triangle ABD,

$\tan(\theta) = \frac{perpendicular}{base}$

$\tan(60) = \frac{AB }{ BD}$

$\tan(60) = \sqrt{3}$

$\sqrt{3} = \frac{h}{x}$

$x = \frac{h}{ \sqrt{3} }$

Substituting the values of x that we got:

$h - 50 = \frac{h}{ \sqrt{3} }$

Cross multiplying :

$\sqrt{3} h - 50 \sqrt{3} = h$

Rearranging the terms:

$\sqrt{3h} - h = 50 \sqrt{3}$

Taking h common:

$h( \sqrt{3} - 1) = 50 \sqrt{3}$

$h = \frac{50 \sqrt{3} }{ \sqrt{3} - 1 }$

Rationalising the term =

$h = \frac{50 \sqrt{3} }{ \sqrt{3} - 1 } \times \frac{ \sqrt{3 } + 1 }{ \sqrt{3} + 1}$

$h = \frac{50 \sqrt{3} ( \sqrt{3} + 1) }{3 - 1}$

$h = \frac{50 (3 + \sqrt{3}) }{2}$

$h = \frac{150+ 50 \sqrt{3} }{2}$

$h = 75 + 25 \sqrt{3}$

h = 75 + 25×1.732

h = 75+43.3

h = 118.3 metres

The height of the tower is 118.3 metres

We got an equation that x = h - 50

X = 118.3-50

X = 68.3 metres

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){4.4}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf50 m}\put(7.3,2){\sf50 m}\put(11.1,4){\sf x m}\qbezier(8,1)(10.7,5)(11,5.4)\qbezier(8,3)(10.5,5)(11,5.4)\qbezier(8.65,1)(8.7,1.4)(8.4,1.6)\put(8.7,1.2){\sf60^{\circ} }\qbezier(8.5,3.4)(8.8,3.2)(8.7,3)\put(8.8,3.2){\sf45^{\circ} }\put(7.7,3){\large{Y}}\put(7.7,1){\large{X}}\put(11.1,1){\large{P}}\put(11.1,3){\large{R}}\put(11.1,5.3){\large{Q}}\put(9.5,0.8){\sf y m}\end{picture}$

$\rule{100}{0.8}$

☢ Given : The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively.

⠀

• A Building ( YX) of height 50 m
• A Tower ( QP) of height (x + 50) m
• Horizontal Distance between Tower and Building be XP i.e. y

⠀

☯ $\underline{\textsf{In \triangle QRY, where \measuredangle R is 90 ^{\circ} :}}$

$:\implies\sf \tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf \tan(45^{\circ})=\dfrac{QR}{YR}\\\\\\:\implies\sf \tan(45^{\circ})=\dfrac{QR}{XP}\qquad[\because\:YR=XP=y]\\\\\\:\implies\sf 1 = \dfrac{x}{y}\\\\\\:\implies\sf y = x \qquad...\:eq \:(l)$

⠀⠀⠀$\rule{160}{1.5}$

☯ $\underline{\textsf{In \triangle QPX, where \measuredangle P is 90 ^{\circ} :}}$

$:\implies\sf \tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf \tan(60^{\circ})=\dfrac{QP}{XP}\\\\\\:\implies\sf \sqrt{3} = \dfrac{x + 50}{y}\\\\\\:\implies\sf \sqrt{3} = \dfrac{x + 50}{x}\qquad[\because\:y=x\:from\:eq.\:l]\\\\\\:\implies\sf \sqrt{3}x = x + 50\\\\\\:\implies\sf \sqrt{3}x - x = 50\\\\\\:\implies\sf x( \sqrt{3} - 1) = 50\\\\\\:\implies\sf x = \dfrac{50}{\sqrt{3} - 1}\\\\\\:\implies\sf x =\dfrac{50}{\sqrt{3} - 1} \times\dfrac{ \sqrt{3} + 1 }{\sqrt{3} + 1}\\\\\\:\implies\sf x = \dfrac{50( \sqrt{3} + 1)}{3 - 1}\\\\\\:\implies\sf x = \dfrac{50(1.73 + 1)}{2}\\\\\\:\implies\sf x = 25 \times 2.73\\\\\\:\implies\sf x = 68.25\:m = y \qquad \bigg\lgroup\bf Distance\:in\:Between\bigg\rgroup$

$\rule{200}{2}$

☯ $\underline{\textsf{Height of Tower :}}$

$\dashrightarrow\sf\:\:Tower=QP\\\\\\\dashrightarrow\sf\:\:Tower=x+50\\\\\\\dashrightarrow\sf\:\:Tower=68.25\:m+50\\\\\\\dashrightarrow\sf\:\:Tower=118.25\:m\qquad \bigg\lgroup\bf Height\:of\:Tower\bigg\rgroup$