Height and Distance

$$SR = PQ = 2\,m$$
$$PS = QR = 10\sqrt 3 \,m$$
$$\tan {30^ \circ } = \frac{{TS}}{{PS}}$$
$$\frac{1}{{\sqrt 3 }} = \frac{{TS}}{{10\sqrt 3 }}$$
$$TS = \frac{{10\sqrt 3 }}{{\sqrt 3 }} = 10\,m$$
$$TR = TS + SR = 10 + 2 = 12\,m$$

Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation
Given that ∠ ACB = 22.5° and ∠ DCE = 67.5°
Given that C is the midpoint of BD
Hence, BC = CD
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABC,$$
$$\tan{22.5^ \circ } = \frac{{AB}}{{BC}}\,.....\left( {eq:1} \right)\,$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,CDE,$$
$$\tan {67.5^ \circ } = \frac{{ED}}{{CD}}\,.....\left( {eq:2} \right)$$
$$\frac{{\left( {eq:2} \right)}}{{\left( {eq:1} \right)}} ⇒ \frac{{\tan {{67.5}^ \circ }}}{{\tan{{22.5}^ \circ }}} = \frac{{\left( {\frac{{ED}}{{CD}}} \right)}}{{\left( {\frac{{AB}}{{BC}}\,} \right)}}$$
$$ = \frac{{ED}}{{AB}}\,\,\,\,\,\left( {\because CD = BC} \right)$$
$$ ⇒ \frac{{\tan \left( {{{90}^ \circ } - {{22.5}^ \circ }} \right)}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}}$$
$$ ⇒ \frac{{\cot {{22.5}^ \circ }}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}}$$
$$\,\,\,\,\,\,\,\,\,\,\left[ {\because \tan \left( {90 - \theta } \right) = \cot \theta } \right]$$
$$ ⇒ \frac{{\left( {\frac{1}{{\tan {{22.5}^ \circ }}}} \right)}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \cot \theta = \frac{1}{{\tan \theta }}} \right]$$
$$ ⇒ \frac{{ED}}{{AB}} = \frac{1}{{{{\left( {\tan {{22.5}^ \circ }} \right)}^2}}}$$
$$ = \frac{1}{{{{\left( {\sqrt 2 - 1} \right)}^2}}}$$
$$ = {\left( {\frac{1}{{\sqrt 2 - 1}}} \right)^2}$$
$$ = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}} \right]^2}$$
$$ = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {2 - 1} \right)}}} \right]^2}$$
$$ = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{1}} \right]^2}$$
$$ = {\left( {\sqrt 2 + 1} \right)^2}$$
$$ = \left( {2 + 2\sqrt 2 + 1} \right)$$
$$ = \left( {3 + 2\sqrt 2 } \right)$$
$${\text{Required}}\,{\text{Ratio}}$$
$$ = ED:AB$$
$$ = \left( {3 + 2\sqrt 2 } \right):1$$

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, ∠ ABC = 60°, ∠ DBC = 30°
Let DC be h
$$\tan {30^ \circ } = \frac{{DC}}{{BC}}$$
$$\frac{1}{{\sqrt 3 }} = \frac{h}{{BC}}$$
$$h = \frac{{BC}}{{\sqrt 3 }}\,......\left( eq : 1 \right)$$
$$\tan {60^ \circ } = \frac{{AC}}{{BC}}$$
$$\sqrt 3 = \frac{{18 + h}}{{BC}}$$
$$18 + h = BC \times \sqrt 3 \,......\left( eq: 2 \right)$$
$$\frac{eq : 1}{eq : 2} ⇒ \frac{h}{{18 + h}} = \frac{{\left( {\frac{{BC}}{{\sqrt 3 }}} \right)}}{{\left( {BC \times \sqrt 3 } \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3}$$
$$ ⇒ 3h = 18 + h$$
$$ ⇒ 2h = 18$$
$$ ⇒ h = 9\,{\text{m}}$$
i.e., the height of the tower = 9 m

Let AB be the man and CD be the window
Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
∠ DAF = 45°, ∠ CAF = 60°
From the diagram, AF = BE = 5 m
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFD,$$
$$\tan {45^ \circ } = \frac{{DF}}{{AF}}$$
$$1 = \frac{{DF}}{5}$$
$$DF = 5\,......\left( 1 \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFC,$$
$$\tan {60^ \circ } = \frac{{CF}}{{AF}}$$
$$\sqrt 3 = \frac{{CF}}{5}$$
$$CF = 5\sqrt 3 \,......\,\left( 2 \right)$$
$${\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{window}}$$
$$ = CD = \left( {CF - DF} \right)$$
  $$ = 5\sqrt 3 - 5$$     [∵ Substituted the value of CF and DF from (1) and (2)]
$$ = 5\left( {\sqrt 3 - 1} \right)$$
$$ = 5\left( {1.73 - 1} \right)$$
$$ = 5 \times 0.73$$
$$ = 3.65\,{\text{m}}$$

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m, ∠ ABD = 30°, ∠ ACD = 60°,
Let CD = x, AD = h
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,CDA$$
$$\tan {60^ \circ } = \frac{{AD}}{{CD}}$$
$$\sqrt 3 = \frac{h}{x}\,......\left( {eq:1} \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,BDA$$
$$\tan {30^ \circ } = \frac{{AD}}{{BD}}$$
$$\frac{1}{{\sqrt 3 }} = \frac{h}{{70 + x}}\,......\left( {eq:2} \right)$$
$$\frac{{eq:1}}{{eq:2}} ⇒ \frac{{\sqrt 3 }}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}} = \frac{{\left( {\frac{h}{x}} \right)}}{{\left( {\frac{h}{{70 + x}}} \right)}}$$
$$ ⇒ 3 = \frac{{70 + x}}{x}$$
$$ ⇒ 2x = 70$$
$$ ⇒ x = 35$$
Substituting this value of x in eq : 1, we have
$$\sqrt 3 = \frac{h}{{35}}$$
$$ ⇒ h = 35\sqrt 3 = 35 \times 1.73$$
$$ = 60.55 \approx 60.6$$

Let AB be the tower of height 150 m
C is car and angle of depression is 30°

∴ ∠ACB = 30° (alternate angle)
In right - angled triangle ABC,
$$\frac{{BC}}{{AB}} = \cot {30^ \circ }$$
$$ ⇒ \frac{{BC}}{{150}} = \sqrt 3 $$
$$ ⇒ BC = 150\sqrt 3 \,m$$
That is, distance of the car from the tower is $$150\sqrt 3 \,m$$

Let AB be tower and a point P distance of 30 m from it's foot of the tower which form an angle of elevation pf the sun of 60°

$${\text{Let height of tower }}AB = h$$
$${\text{Then in right }}\Delta APB,$$
$$\tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}}$$
$$ ⇒ \tan {60^ \circ } = \frac{h}{{30}}$$
$$ ⇒ \sqrt 3 = \frac{h}{{30}}$$
$$ ⇒ h = 30\sqrt 3 $$
∴ Height of the tower $$ = 30\sqrt 3 \,m$$

Let AB and CD are two poles AB = 10 m and CD = 16 m

AC is wire which makes an angle of 30° with the horizontal
Let BD = x, then AE = x
CE = CD - ED = CD - AB = 16 - 10 = 6m
$${\text{Now}}\,{\text{in}}\,\Delta ACE$$
$$\sin {30^ \circ } = \frac{{CE}}{{AC}} = \frac{6}{l}$$
$$ ⇒ \frac{1}{2} = \frac{6}{l}$$
$$ ⇒ l = 2 \times 6 = 12\,m$$

Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y

$${\text{Now in right }}\Delta ADB,$$
$$\tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{DB}}$$
$$\tan {45^ \circ } = \frac{{100}}{y}$$
$$ ⇒ 1 = \frac{{100}}{y}$$
$$ ⇒ y = 100 $$
$$ {\text{Similarly in right }}\Delta ACB,$$
$$\tan {30^ \circ } = \frac{{AB}}{{CB}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{{100}}{{y + x}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{{100}}{{100 + x}}$$
$$ ⇒ 100 +×= 100\sqrt 3 $$
$$ ⇒ x = 100\sqrt 3 - 100$$
$$ ⇒ x = 100\left( {\sqrt 3 - 1} \right)\,m$$

Let AB be the tower and C is a point such that the angle of elevation of A is α.
After walking towards the foot B of the tower, at D the angle of elevation is β.
Let h be the height of the tower and DB =×Now in ΔACB,

$$\tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{CB}}$$
$$\tan \alpha = \frac{h}{{d + x}}$$
$$ ⇒ d +×= \frac{h}{{\tan \alpha }}$$
$$ ⇒ d +×= h\cot \alpha $$
$$ ⇒ x = h\cot \alpha - d\,.......\left( {\text{i}} \right)$$
$${\text{Similarly in right }}\Delta ADB,$$
$$\tan \beta = \frac{h}{x}$$
$$ ⇒ x = \frac{h}{{\tan \beta }}$$
$$ ⇒ x = h\cot \beta \,.........({\text{ii}})$$
$${\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right)$$
$$h\cot \alpha - d = h\cot \beta $$
$$ ⇒ h\cot \alpha - h\cot \beta = d$$
$$ ⇒ h\left( {\cot \alpha - \cot \beta } \right) = d$$
$$ ⇒ h = \frac{d}{{\cot \alpha - \cot \beta }}$$
∴ Height of the tower $$ = \frac{d}{{\cot \alpha - \cot \beta }}$$