Height And Distance — Part 1 MCQs with Explanations

Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45° $$\frac{{AB}}{{AC}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$ $$ \Rightarrow AC = AB \times \sqrt 3 =100 \sqrt 3{\text{m}}$$ $$\frac{{AB}}{{AD}} = \tan {45^ \circ } = 1$$ $$ \Rightarrow AD = AB = 100{\text{m}}$$ ∴ CD = (AC+AD) = (100√3 +100) = 100(√3 +1) = 100(1.73+1) =100 × 2.73 = 273m

One of AB, AD and CD must have given. So, the data is inadequate.

Let AB be the wall and BC be the ladder. Then, ∠ACB = 60° = AC = 4.6m $$\frac{{AC}}{{BC}} = \cos {60^ \circ } = \frac{1}{2}$$ ⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

Let AB be the observer and CD be the tower. Draw BE ⊥ CD Then CE = AB = 1.6m BE = AC = 20√3m $$\frac{{DE}}{{BE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$ $$ \Rightarrow DE = \frac{{BE}}{{\sqrt 3 }} = \frac{{20\sqrt 3 }}{{\sqrt 3 }} = 20$$ ∴ CD = CE + DE = (1.6 + 20) m = 21.6 m

Let AB be the tower. Then, ∠APB = 30° and AB = 100m $$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$ ⇒ AP = AB × √3 = 100 × √3 ⇒ AP = 100 × 1.73 = 173m

Let AB be the tree and AC be its 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, $$\eqalign{ & \frac{{AB}}{{AC}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AC = AB \times \sqrt 3 = h\sqrt 3 {\kern 1pt} {\text{And,}} \cr & \Rightarrow \frac{{AB}}{{AD}} = \tan 45^\circ = 1 \cr & \Rightarrow AD = AB = h \cr & \, \, \, \, \, CD = 20 \cr & \Rightarrow \left( {AC - AD} \right) = 20 \cr & \Rightarrow h\sqrt 3 - h = 20 \cr & \therefore h = \frac{{20}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = 10\left( {\sqrt 3 + 1} \right){\text{m}} \cr & = \left( {10 \times 2.73} \right){\text{m}} \cr & = 27.3 {\text{m}} \cr} $$

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

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

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 = x and AC = y $$\eqalign{ & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow \frac{h}{y} = 1 \cr & \Rightarrow y = h \cr & \frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x + y = \sqrt 3 h \cr & \therefore x = \left( {x + y - y} \right) \cr & \,\,\,\,\,\,\, = \sqrt 3 h - h \cr & \,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right) \cr} $$ Now, $$h\left( {\sqrt 3 - 1} \right)$$ is covered in 12 min. So, h will be covered in→ $$\eqalign{ & \left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right] \cr & = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min \cr & = \left( {\frac{{1200}}{{73}}} \right)\min \cr & = 16\min ,\,\,23\sec \cr} $$

$$\eqalign{ & {\text{AC = 30 meter}} \cr & {\text{AB = 15 meter}} \cr & \angle {\text{ACB}} = \theta \cr & \therefore \sin \theta = \frac{{AB}}{{AC}} = \frac{{15}}{{30}} = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr} $$

$$\eqalign{ & {\text{tan }}\theta = \frac{H}{{{m^2}}} \cr & \Rightarrow \tan \left( {90^ \circ - \theta } \right) = \frac{H}{{{n^2}}} \cr & \Rightarrow \cot \theta = \frac{H}{{{n^2}}} \cr & \Rightarrow \tan \theta .\cot \theta = \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} \cr & \Rightarrow \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} = 1 \cr & \Rightarrow {H^2} = {m^2}{n^2} \cr & \Rightarrow H = mn \cr} $$

$$\eqalign{ & {\text{Here, AD is parallel to BC and AC is a transversal}}{\text{.}} \cr & \Rightarrow {\theta _1} = {\theta _2} \cr} $$

$$\eqalign{ & {\text{Let, }}AC = H{\text{ }}m \cr & OA = {\text{ }}a{\text{ }}m \cr & \tan {\text{ }}{30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{H}{a} \cr} $$ $$\tan {60^ \circ } = \sqrt 3 $$ $$ = \frac{{H + 200 + 200}}{a}$$ $$\eqalign{ & \Rightarrow \frac{{H + 400}}{{\sqrt 3 H}} = \sqrt 3 \cr & \Rightarrow H + 400 = 3H \cr & \Rightarrow 2H = 400\,m \cr & \Rightarrow H = 200\,m \cr} $$ Height of cloud above lake = 200 m

Let AB be the tower and C and D be the two positions of the boats. Then, $$\angle {\text{ACB = }}{45^ \circ },$$ $$\angle ADB = {30^ \circ }$$ and AC = 60 m Let, AB = h $$\eqalign{ & {\text{Then,}}\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AB = AC \cr & \Rightarrow h = 60\,m \cr & {\text{And}}\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AD = \left( {AB \times \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 60\sqrt 3 \,m \cr & \therefore CD = \left( {AD - AC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 60\left( {\sqrt 3 - 1} \right)\,m \cr & {\text{Hence, required speed }} \cr & {\text{ = }}\left[ {\frac{{60\left( {\sqrt 3 - 1} \right)}}{5}} \right]{\text{m/s}} \cr & {\text{ = }}\left( {12 \times 0.73} \right){\text{m/s}} \cr & = \left( {12 \times 0.73 \times \frac{{18}}{5}} \right){\text{km/hr}} \cr & = {\text{31}}{\text{.5km/hr }} \approx {\text{ 32km/hr}} \cr} $$

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) $$\eqalign{ & \frac{{AB}}{{AC}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AC = \frac{{AB}}{{\sqrt 3 }} = \frac{{15}}{{\sqrt 3 }} \cr & {\text{And, }}\frac{{BE}}{{DE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow DE = \left( {BE \times \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {15 - h} \right) \cr & {\text{So, }}AC = DE \cr & \Rightarrow \frac{{15}}{{\sqrt 3 }} = \sqrt 3 \left( {15 - h} \right) \cr & \Rightarrow 3h = \left( {45 - 15} \right) \cr & \Rightarrow h = 10{\text{ m}} \cr} $$

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

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

Let AB be lighthouse and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h Let PB = x and QB = y $$\eqalign{ & {\text{Now in right }}\Delta APB \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {45^ \circ } = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow x = h\,............(i) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan {30^ \circ } = \frac{{AP}}{{QB}} = \frac{h}{y} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{y} \cr & \Rightarrow y = \sqrt 3 \,h\,..............(ii) \cr & {\text{Adding (i) and (ii)}} \cr & \therefore PQ = x + y \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = h + \sqrt 3 \,h \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt 3 + 1} \right)\,h \cr} $$

Let AB is girls and CD is lamp-post AB = 1.5 which casts her shadow EB $$\eqalign{ & \therefore EB = 4.5\,m,\,BD = 3\,m \cr & {\text{Now in }}\Delta AEB \cr & \tan \theta = \frac{{AB}}{{BE}} = \frac{{1.5}}{{4.5}} = \frac{1}{3} \cr & {\text{and in }}\Delta CED \cr & \tan \theta = \frac{{CD}}{{ED}} \Rightarrow \frac{1}{3} = \frac{h}{{4.5 + 3}} \cr & \Rightarrow \frac{1}{3} = \frac{h}{{7.5}} \cr & \Rightarrow h = \frac{{7.5}}{3} - 2.5\,m \cr & \therefore {\text{Height of lamp - post}} = {\text{2}}{\text{.5}}\,m \cr} $$

Let AB be a vertical pole and let its shadow be BC Let BC = x m, then length of pole = $$\sqrt 3 $$ x, $$\theta $$ be the angle of elevation $$\eqalign{ & \therefore {\text{tan}}\theta = \frac{{AB}}{{BC}} = \frac{{\sqrt 3 \,x}}{x} = \sqrt 3 \cr & = \tan {60^ \circ } \cr & \therefore \theta = {60^ \circ } \cr} $$

Let AB and CD are two persons standing ‘a’ meters apart P is the mid-point of BD and from M, the angles of elevation of A and C are complementary $$\eqalign{ & {\text{In}}\,\,\Delta {\text{APB,}} \cr & \tan \theta = \frac{{AB}}{{BP}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a} \cr & {\text{In}}\,\,\Delta {\text{CDP,}} \cr & \cot (90 - \theta ) = \frac{{PD}}{{CD}} = \frac{{\frac{a}{2}}}{{2h}} = \frac{a}{{4h}} \cr & {\text{We}}\,\,{\text{Know}}\,\,{\text{that,}} \cr & \tan \theta = \cot (90 - \theta ). \cr & \therefore \frac{{2h}}{a} = \frac{a}{{4h}} \cr & \Rightarrow 8{h^2} = {a^2} \cr & \Rightarrow h = \frac{a}{{2\sqrt 2 }} \cr} $$

Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60° Let height of the tower = h $$\eqalign{ & {\text{Then in right }}\Delta ABP \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{100}} \Rightarrow \sqrt 3 = \frac{h}{{100}} \cr & \Rightarrow h = 100\sqrt 3 \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = 100\sqrt 3 \cr} $$

Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively ∠ACB = ∠XAC = 30° , ∠ADB = ∠YAD = 45° and CD = 100m Let AB =h and CB = x then BC = (100 - x)m $$\eqalign{ & {\text{Now in }}\,\Delta ACB, \cr & \tan \theta = \frac{{AB}}{{CB}} \cr & \tan {30^ \circ } = \frac{h}{x} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 h ......{\text{(i)}} \cr & {\text{Similarly in }}\,\Delta ADB \cr & \tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & \Rightarrow 1 = \frac{h}{{100 - x}} \cr & \Rightarrow x = 100 - h \,......{\text{(ii)}} \cr & {\text{From (i) and (ii)}} \cr & \sqrt 3 h = 100 - h \cr & \Rightarrow (\sqrt 3 - 1)h = 100 \cr & h = \frac{{100}}{{\sqrt 3 + 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{2} \cr & \,\,\,\,\,\, = 50\left( {\sqrt 3 - 1} \right) \cr} $$ ∴ height of light house $$ = 50\left( {\sqrt 3 - 1} \right)m$$

AB is a tower and AB = 75 m From A, the angle of depression of a car C on the ground is 30° $$\eqalign{ & {\text{Let distance }}BC = x \cr & {\text{Now in right }}\Delta ACB, \cr & \tan \theta = \frac{{AB}}{{BC}} \cr & \Rightarrow \tan {30^ \circ } = \frac{{75}}{x} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{75}}{x} \cr & \Rightarrow x = 75\sqrt 3 \,m \cr & \therefore BC = 75\sqrt 3 \,m \cr} $$

Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively $$\eqalign{ & {\text{Let }}AB = h \cr & {\text{Now in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {30^ \circ } = \frac{h}{a} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{a}\,...........(i) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan {60^ \circ } = \frac{{AB}}{{QB}} \cr & \Rightarrow \sqrt 3 = \frac{h}{b}\,...........(ii) \cr & {\text{Multiplying (i) and (ii)}} \cr & \frac{1}{{\sqrt 3 }} \times \sqrt 3 = \frac{h}{a} \times \frac{h}{b} \cr & \Rightarrow 1 = \frac{{{h^2}}}{{ab}} \cr & \Rightarrow {h^2} = ab \cr & \Rightarrow h = \sqrt {ab} \cr & \therefore {\text{Height of the tower}} = \sqrt {ab} \cr} $$

Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground ∴ In ∆ABC, ∠B = 90° Let height of wall AB = h Then $$\sin \theta = \frac{{AB}}{{AC}} \Rightarrow \sin {60^ \circ } = \frac{h}{{15}}$$ $$\eqalign{ & \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{h}{{15}} \cr & \Rightarrow h = \frac{{15\sqrt 3 }}{2}\,m \cr} $$ ∴ Height of the wall $$ = \frac{{15\sqrt 3 }}{2}\,m$$

Let height of pole CD = h and AB = 2h, BD = a M is mid-point of BD $$\therefore DM = MB = \frac{a}{2}$$ $${\text{Let }}\angle CMD = \theta ,$$ $${\text{then }}\angle AMB = $$ $${90^ \circ } - \theta $$ $$\eqalign{ & {\text{Now}} \cr & \tan \theta = \frac{{CD}}{{DM}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a}\,......({\text{i}}) \cr & {\text{and}} \cr & tan\left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{MB}} = \frac{{2h}}{{\frac{a}{2}}} = \frac{{4h}}{a} \cr & \Rightarrow \cot \theta = \frac{{4h}}{a}\,..........({\text{ii}}) \cr & {\text{Multiplying (i) and (ii)}} \cr & {\text{tan}}\theta \times {\text{cot}}\theta = \frac{{2h}}{a} \times \frac{{4h}}{a} \cr & 1 = \frac{{8{h^2}}}{{{a^2}}} = {h^2} = \frac{{{a^2}}}{8}\,m \cr & h = \sqrt {\frac{{{a^2}}}{8}} = \frac{a}{{\sqrt 8 }} = \frac{a}{{2\sqrt 2 }}\,m \cr} $$

Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C Let angle be Q and CD = 25 m $$\eqalign{ & {\text{Let}}\,AB = h \cr & CE \,\, || \,\, DB \cr & \therefore EC = DB = x{\text{ }}\left( {{\text{suppose}}} \right) \cr & EB = CD = 25 \cr & \therefore AE = h - 25 \cr & {\text{Now in right }}\Delta CDB, \cr & \tan \theta = \frac{{CD}}{{DB}} = \frac{{25}}{x}\,......\left( {\text{i}} \right) \cr & {\text{and in right }}\Delta CAE \cr & \tan \theta = \frac{{AE}}{{CE}} = \frac{{h - 25}}{x}\,......\left( {{\text{ii}}} \right) \cr & {\text{From}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{25}}{x} = \frac{{h - 25}}{x} \cr & \Rightarrow 25 = h - 25 \cr & \Rightarrow h = 25 + 25 = 50 \cr & \therefore {\text{Height of tower}} = 50\,m \cr} $$

Let AB be tower and C is a point on the ground 50 m away From foot of tower B Angle of elevation is 45° Let h be height of tower = x m $$\eqalign{ & \therefore \tan \theta = \frac{{AB}}{{BC}} \cr & \Rightarrow \tan {45^ \circ } = \frac{{AB}}{50} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \Rightarrow AB = 50\,m \cr} $$

Let AB be rod and BC be its shadow So that AB : BC = 1 : $$\sqrt 3 $$ Let $$\theta $$ be the angle of elevation $$\eqalign{ & \therefore \tan \theta = \frac{{AB}}{{BC}} = \frac{1}{{\sqrt 3 }} = \tan {30^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \tan {{30}^ \circ } = \frac{1}{{\sqrt 3 }}} \right) \cr & \therefore \theta = {30^ \circ } \cr} $$ ∴ Hence angle of elevation $$ = {30^ \circ }$$

In the figure, AB is chimney and CB and DB are its shadow $$\eqalign{ & \tan {60^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{h}{{BC}} \cr & \Rightarrow BC = \frac{h}{{\sqrt 3 }}\,.......\,\left( {\text{i}} \right) \cr & {\text{and}} \cr & \tan {30^ \circ } = \frac{h}{{DB}} = \frac{h}{{DB + BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{x + BC}} \cr & x + BC = h\sqrt 3 \cr & \Rightarrow BC = h\sqrt 3 - x\,.......\,\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & \frac{h}{{\sqrt 3 }} = h\sqrt 3 - x \cr & \Rightarrow \frac{h}{{\sqrt 3 }} - h\sqrt 3 = - x \cr & x = h\sqrt 3 - \frac{h}{{\sqrt 3 }} \cr & x = h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) \cr & x = h\frac{{3 - 1}}{{\sqrt 3 }} \cr & x = \frac{{2h}}{{\sqrt 3 }} \cr & \therefore h = \frac{{\sqrt 3 }}{2}x \cr} $$

Suppose AB is the ladder of length x m ∴ OA = 2m, ∠OAB = 60° $$\eqalign{ & {\text{In right }}\Delta AOB,\,\sec {60^ \circ } = \frac{x}{2} \cr & \Rightarrow 2 = \frac{x}{2} \cr & \Rightarrow x = 4\,m \cr} $$

Let AB and CD be two poles AB = 20 m, CD = 14 m A and C are joined by a wire CE || DB and angle of elevation of A is 30° Let CE = DB = x and AC = L Now AE = AB - EB = AB - CD = 20 - 14 = 6 m $$\eqalign{ & {\text{Now in right }}\Delta ACE, \cr & \sin \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} = \frac{{AE}}{{AC}} \cr & \Rightarrow \sin {30^ \circ } = \frac{6}{{AC}} \cr & \Rightarrow \frac{1}{2} = \frac{6}{{AC}} \cr & \Rightarrow AC = 2 \times 6 = 12 \cr & \therefore {\text{Length of AC}} = 12\,m \cr} $$

Let CD is the tower and A is a point such that the angle of elevation of C is 30° B is and their point h m high of A and angle of depression of D is 60° $$\eqalign{ & {\text{The}}\,AB = h\,m \cr & {\text{Let}}\,CD = H\,m\,\,\,{\text{and }}\,AD\, = x \cr & {\text{Now}}\,{\text{in}}\,{\text{right}}\,\Delta ABD, \cr & \tan \theta = \frac{{AB}}{{AD}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{x} \cr & \Rightarrow \sqrt 3 = \frac{h}{x} \cr & \Rightarrow x = \frac{h}{{\sqrt 3 }}\,.......({\text{i}}) \cr & {\text{Similarly in right }}\Delta ACB, \cr & \tan {30^ \circ } = \frac{{CD}}{{AD}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{H}{x} \cr & \Rightarrow x = \sqrt 3 \,H\,.......\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & \sqrt 3 \,H = \frac{h}{{\sqrt 3 }} \cr & H = \frac{h}{{\sqrt 3 \times \sqrt 3 }} = \frac{h}{3} \cr & \therefore {\text{Height of tower}} = \frac{h}{3} \cr} $$

Let AB be the tower of height 150 m C is car and angle of depression is 30° Therefore, ∠ACB = 30° (alternate angle) In right - angled triangle ABC, $$\eqalign{ & \frac{{BC}}{{AB}} = \cot {30^ \circ } \cr & \Rightarrow \frac{{BC}}{{150}} = \sqrt 3 \cr & \Rightarrow BC = 150\sqrt 3 \,m \cr} $$ 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 its foot of the tower which form an angle of elevation pf the sun of 60° $$\eqalign{ & {\text{Let height of tower }}AB = h \cr & {\text{Then in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{30}} \cr & \Rightarrow \sqrt 3 = \frac{h}{{30}} \cr & \Rightarrow h = 30\sqrt 3 \cr} $$ ∴ 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 $$\eqalign{ & {\text{Now}}\,{\text{in}}\,\Delta ACE \cr & \sin {30^ \circ } = \frac{{CE}}{{AC}} = \frac{6}{l} \cr & \Rightarrow \frac{1}{2} = \frac{6}{l} \cr & \Rightarrow l = 2 \times 6 = 12\,m \cr} $$

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

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