Height and Distance

Let AB be tower and BC be it's shadow
∴ Let AB = x

$${\text{Then}}\,BC = \sqrt 3 \times×= \sqrt 3 \,x$$
$$\therefore \tan \theta = \frac{{AB}}{{BC}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{{\sqrt 3 \,x}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \tan {30^ \circ }$$
$$\therefore \theta = {30^ \circ }$$
∴ Angle of elevation of the sun$${\text{ = }}{30^ \circ }$$

AB is a tower
BD and BC are it's shadows and CD = 2x

$$\tan {45^ \circ } = \frac{{AB}}{{DB}}$$
$$ ⇒ 1 = \frac{h}{y} ⇒ y = h$$
$${\text{and}}\tan {30^ \circ } = \frac{{AB}}{{CB}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{h}{{2x + y}}$$
$$ ⇒ 2x + y = \sqrt 3 \,h$$
$$ ⇒ \sqrt 3 \,h - h = 2x$$
$$ ⇒ h\left( {\sqrt 3 - 1} \right) = 2x$$
$$ ⇒ h = \frac{{2x}}{{\sqrt 3 - 1}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{3 - 1}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {\sqrt 3 + 1} \right)$$

Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are $$\theta $$ and (90° – $$\theta $$)
Let height of tower = h

$${\text{Then}}\,{\text{in}}\,{\text{right}}\,\Delta APB,$$
$$\tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}}$$
$$ = \frac{h}{a}\,............\left( {\text{i}} \right)$$
$${\text{Similarly in right }}\Delta AQB,$$
$$\tan \left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{QB}} = \frac{h}{b}$$
$$ ⇒ \cot \theta = \frac{h}{b}\,...........\left( {{\text{ii}}} \right)$$
$${\text{Multiplying }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right)$$
$$\tan \theta \cot \theta = \frac{h}{a} \times \frac{h}{b}$$
$$ ⇒ 1 = \frac{{{h^2}}}{{ab}}$$
$$ ⇒ {h^2} = ab$$
$$ ⇒ h = \sqrt {ab} $$
$$\therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = \sqrt {ab} $$

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°, ∠BCD = 60°
$$\tan {45^ \circ } = \frac{{BD}}{{BA}}$$
$$ ⇒ 1 = \frac{{60}}{{BA}}$$
$$ ⇒ BA = 60\,m\,........\left( {\text{i}} \right)$$
$$\tan {60^ \circ } = \frac{{BD}}{{BC}}$$
$$ ⇒ \sqrt 3 = \frac{{60}}{{BC}}$$
$$ ⇒ BC = \frac{{60}}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60\sqrt 3 }}{3}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\sqrt 3 $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 1.73$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.6\,m\,..........\left( {{\text{ii}}} \right)$$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m

Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m, ∠DAC = 45°, ∠DBC = 60°
$$\tan {60^ \circ } = \frac{{DC}}{{BC}}$$
$$\sqrt 3 = \frac{{600}}{{BC}}$$
$$BC = \frac{{600}}{{\sqrt 3 }}\,..........\,\left( {\text{i}} \right)$$
$$\tan {45^ \circ } = \frac{{DC}}{{AC}}$$
$$1 = \frac{{600}}{{AC}}$$
$$AC = 600\,............\,\left( {{\text{ii}}} \right)$$
Distance between the objects
$$ = AC = \left( {AC - BC} \right)$$
$$ = 600 - \frac{{600}}{{\sqrt 3 }}$$   [∵ from (1) and (ii)]
$$ = 600\left( {1 - \frac{1}{{\sqrt 3 }}} \right)$$
$$ = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right)$$
$$ = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \times \frac{{\sqrt 3 }}{{\sqrt 3 }}$$
$$ = \frac{{600\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{3}$$
$$ = 200\sqrt 3 \left( {\sqrt 3 - 1} \right)$$
$$ = 200\left( {3 - \sqrt 3 } \right)$$
$$ = 200\left( {3 - 1.73} \right)$$
$$ = 254\,m$$

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 200 m, ∠BAD = 30°, ∠BCD = 45°
$$\tan {30^ \circ } = \frac{{BD}}{{BA}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{{200}}{{BA}}$$
$$ ⇒ BA = 200\sqrt 3 $$
$$\tan {45^ \circ } = \frac{{BD}}{{BC}}$$
$$ ⇒ 1 = \frac{{200}}{{BC}}$$
$$ ⇒ BC = 200$$
Distance between the two ships
$$ = AC = BA + BC$$
$$ = 200\sqrt 3 + 200$$
$$ = 200(\sqrt 3 + 1)$$
$$ = 200\left( {1.73 + 1} \right)$$
$$ = 200 \times 2.73$$
$$ = 546{\text{ }}m$$

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have ∠DAC = 30°, ∠DBC = 45°
Let DC = h
$$\tan {30^ \circ } = \frac{{DC}}{{AC}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{h}{{AC}}$$
$$ ⇒ AC = h\sqrt 3 \,.........\,\left( {\text{i}} \right)$$
$$\tan {45^ \circ } = \frac{{DC}}{{BC}}$$
$$ ⇒ 1 = \frac{h}{{BC}}$$
$$ ⇒ BC = h\,........\,\left( {{\text{ii}}} \right)$$
$${\text{We know that,}}$$
$$AB = \left( {AC - BC} \right)$$
$$ ⇒ 40 = \left( {AC - BC} \right)$$
$$ ⇒ 40 = \left( {h\sqrt 3 - h} \right)$$   [∵ from (1) and (ii)]
$$ ⇒ 40 = h\left( {\sqrt 3 - 1} \right)$$
$$ ⇒ h = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}}$$
$$ = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}$$
$$ = \frac{{40\left( {\sqrt 3 + 1} \right)}}{{\left( {3 - 1} \right)}}$$
$$ = \frac{{40\left( {\sqrt 3 + 1} \right)}}{2}$$
$$ = 20\left( {\sqrt 3 + 1} \right)$$
$$ = 20\left( {1.73 + 1} \right)$$
$$ = 20 \times 2.73$$
$$ = 54.6\,m$$

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.
$$\cos {60^ \circ } = \frac{{PQ}}{{PR}}$$
$$\frac{1}{2} = \frac{{12.4}}{{PR}}$$
$$PR = 2 \times 12.4$$
$$\,\,\,\,\,\,\,\,\,\, = 24.8\,m$$

Let RQ be the pole and PQ be the shadow
Given that RQ = 18 m and PQ = $$6\sqrt 3 $$ m
Let the angle of elevation, ∠RPQ = $$\theta $$
$${\text{From the right}}\Delta PQR,$$
$${\text{tan}}\,\theta = \frac{{RQ}}{{PQ}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{6\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 $$
$$ ⇒ \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {60^ \circ }$$

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, ∠BCA = 60°
$$\tan {60^ \circ } = \frac{{BA}}{{CA}}$$
$$\sqrt 3 = \frac{{BA}}{{150}}$$
$$BA = 150\sqrt 3 $$
i.e, the distance travelled by the balloon = $$150\sqrt 3 $$   meters
time taken = 2 min = 2 × 60 = 120 seconds
$${\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}}$$
$$ = \frac{{150\sqrt 3 }}{{120}} = 1.25\sqrt 3 $$
= 1.25 × 1.73 = 2.16 meter/second