Height and Distance

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

Let AB be the tree and AC be it's shadow.

Let ∠ACB = θ
$$\frac{{AC}}{{AB}} = \cot \theta $$
$$\frac{{AC}}{{AB}} = \sqrt 3 $$
$$\cot {30^ \circ } = \sqrt 3 $$
∴ θ = 30°

Let AB be the tower and C and D be the points of observation.
Then, ∠ACB = 30°, ∠ADB = 45° and CD = 20m
Let AB = h then,
$$\frac{{AB}}{{AC}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ AC = AB \times \sqrt 3 = h\sqrt 3 \  {\text{And,}}$$
$$ ⇒ \frac{{AB}}{{AD}} = \tan 45^\circ = 1$$
$$ ⇒ AD = AB = h $$
$$ \, \, \, \, \, CD = 20$$
$$ ⇒ \left( {AC - AD} \right) = 20$$
$$ ⇒ h\sqrt 3 - h = 20$$
$$\therefore h = \frac{{20}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}$$
$$ = 10\left( {\sqrt 3 + 1} \right){\text{m}}$$
$$ = \left( {10 \times 2.73} \right){\text{m}}$$
$$ = 27.3 {\text{m}}$$

Let AB be the tower and C and D be the objects.
Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
$$\frac{{AB}}{{AD}} = \tan {60^ \circ } = \sqrt 3 $$
$$ ⇒ AD = \frac{{AB}}{{\sqrt 3 }} = \frac{{150}}{{\sqrt 3 }}$$
$$\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1$$
$$ ⇒ AC = AB = 150{\text{ m}}$$
$$\therefore CD = \left( {AC - AD} \right)$$
$$ = \left( {150 - \frac{{150}}{{\sqrt 3 }}} \right){\text{m}}$$
$$ = \left[ {\frac{{150\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}} \right]{\text{m}}$$
$$ = 50\left( {3 - \sqrt 3 } \right){\text{m}}$$
$$ = \left( {50 \times 1.27} \right){\text{m}}$$
$$ = 63.5\,{\text{m}}$$

Length of the tower AB = h meter.
$$\angle DAC = \angle ACB = {60^ \circ }$$
$$BC = 70{\text{ meter}}$$
$${\text{In }}\vartriangle {\text{ABC,}}$$
$${\text{tan }}{60^ \circ } = \frac{{AB}}{{BC}}$$
$$ ⇒ \sqrt 3 = \frac{h}{{70}}$$
$$ ⇒ h = 70\sqrt 3 {\text{ meter}}$$

Let AB be the tower and C and D be the two positions of the car.
Then, \[\angle ACB = {45^ \circ },\]    \[\angle ADB = {30^ \circ }\]
Let, AB = h, CD =×and AC = y
$$\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1$$
$$ ⇒ \frac{h}{y} = 1$$
$$ ⇒ y = h$$
$$\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ x + y = \sqrt 3 h$$
$$\therefore×= \left( {x + y - y} \right)$$
$$\,\,\,\,\,\,\, = \sqrt 3 h - h$$
$$\,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right)$$
Now, $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 12 min.
So, h will be covered in→
$$\left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right]$$
$$ = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min $$
$$ = \left( {\frac{{1200}}{{73}}} \right)\min $$
$$ = 16\min ,\,\,23\sec $$