Height and Distance

Let AB be the tower and CD be the electric pole.
Then, $$\angle ACB = {60^ \circ },$$     $$\angle EDB = {30^ \circ }$$     and AB = 15 m
Let CD = h
Then, BE = (AB - AE) = (AB - CD) = (15 - h)
$$\frac{{AB}}{{AC}} = \tan {60^ \circ } = \sqrt 3 $$
$$ ⇒ AC = \frac{{AB}}{{\sqrt 3 }} = \frac{{15}}{{\sqrt 3 }}$$
$${\text{And, }}\frac{{BE}}{{DE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ DE = \left( {BE \times \sqrt 3 } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {15 - h} \right)$$
$${\text{So, }}AC = DE$$
$$ ⇒ \frac{{15}}{{\sqrt 3 }} = \sqrt 3 \left( {15 - h} \right)$$
$$ ⇒ 3h = \left( {45 - 15} \right)$$
$$ ⇒ h = 10{\text{ m}}$$

$${\text{Given tan }}{x^ \circ } = \frac{2}{5}$$     and AF = 200 meter
$$\therefore \frac{2}{5} = \frac{{TF}}{{AF}}$$
$$ ⇒ TF = \frac{{2 \times 200}}{5}$$
$$ ⇒ TF = 80{\text{ m}}$$
$${\text{We have, BF = 80 m}}$$
$$\therefore \tan {\text{ }}{y^ \circ } = \frac{{TF}}{{BF}}$$
$$ ⇒ \tan {\text{ }}{y^ \circ } = \frac{{80}}{{80}}$$
$$ ⇒ \tan {\text{ }}{y^ \circ } = 1 = \tan {45^ \circ }$$
$$\therefore {y^ \circ } = {45^ \circ }$$

$${\text{Let, }}OP = {\text{ }}a$$
$${\text{tan }}{60^ \circ } = \frac{H}{a}$$
$$ ⇒ H = \sqrt 3 a$$
$$ ⇒ \frac{H}{{\sqrt 3 }} = a.....(i)$$
$$tan{45^ \circ } = 1$$   $$ = \frac{H}{{a + 100\left( {3 - \sqrt 3 } \right)}}$$
$$ ⇒ a + 100\left( {3 - \sqrt 3 } \right) = H$$
From (i) $$\frac{H}{{\sqrt 3 }} + $$   $$100\left( {3 - \sqrt 3 } \right)$$   = H
$$ ⇒ H + 300\sqrt 3 - 300 = \sqrt 3 H$$
$$ ⇒ 300\sqrt 3 - 300 = \sqrt 3 H - H$$
$$ ⇒ \left( {\sqrt 3 - 1} \right)H = 300\left( {\sqrt 3 - 1} \right)$$
$$ ⇒ H = 300{\text{ m}}$$

Let AB be lighthouse and P and Q are two ships on it's opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB =×and QB = y

$${\text{Now in right }}\Delta APB$$
$$\tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}}$$
$$ ⇒ \tan {45^ \circ } = \frac{h}{x} ⇒ 1 = \frac{h}{x}$$
$$ ⇒ x = h\,............(i)$$
$${\text{Similarly in right }}\Delta AQB,$$
$$\tan {30^ \circ } = \frac{{AP}}{{QB}} = \frac{h}{y}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{h}{y}$$
$$ ⇒ y = \sqrt 3 \,h\,..............(ii)$$
$${\text{Adding (i) and (ii)}}$$
$$\therefore PQ = x + y$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = h + \sqrt 3 \,h$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt 3 + 1} \right)\,h$$

Let AB is girls and CD is lamp-post AB = 1.5
which casts her shadow EB

$$\therefore EB = 4.5\,m,\,BD = 3\,m$$
$${\text{Now in }}\Delta AEB$$
$$\tan \theta = \frac{{AB}}{{BE}} = \frac{{1.5}}{{4.5}} = \frac{1}{3}$$
$${\text{and in }}\Delta CED$$
$$\tan \theta = \frac{{CD}}{{ED}} ⇒ \frac{1}{3} = \frac{h}{{4.5 + 3}}$$
$$ ⇒ \frac{1}{3} = \frac{h}{{7.5}}$$
$$ ⇒ h = \frac{{7.5}}{3} - 2.5\,m$$
$$\therefore {\text{Height of lamp - post}} = {\text{2}}{\text{.5}}\,m$$

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 $$ = \frac{d}{{\cot \alpha - \cot \beta }}$$