Height And Distance — Part 2 MCQs with Explanations

Let AB be tower and BC be its shadow ∴ Let AB = x $$\eqalign{ & {\text{Then}}\,BC = \sqrt 3 \times x = \sqrt 3 \,x \cr & \therefore \tan \theta = \frac{{AB}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{{\sqrt 3 \,x}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \tan {30^ \circ } \cr & \therefore \theta = {30^ \circ } \cr} $$ ∴ Angle of elevation of the sun$${\text{ = }}{30^ \circ }$$

AB is a tower BD and BC are its shadows and CD = 2x $$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{DB}} \cr & \Rightarrow 1 = \frac{h}{y} \Rightarrow y = h \cr & {\text{and}}\tan {30^ \circ } = \frac{{AB}}{{CB}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{2x + y}} \cr & \Rightarrow 2x + y = \sqrt 3 \,h \cr & \Rightarrow \sqrt 3 \,h - h = 2x \cr & \Rightarrow h\left( {\sqrt 3 - 1} \right) = 2x \cr & \Rightarrow h = \frac{{2x}}{{\sqrt 3 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {\sqrt 3 + 1} \right) \cr} $$

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

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° $$\eqalign{ & \tan {45^ \circ } = \frac{{BD}}{{BA}} \cr & \Rightarrow 1 = \frac{{60}}{{BA}} \cr & \Rightarrow BA = 60\,m\,........\left( {\text{i}} \right) \cr & \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{{60}}{{BC}} \cr & \Rightarrow BC = \frac{{60}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60\sqrt 3 }}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.6\,m\,..........\left( {{\text{ii}}} \right) \cr} $$ 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° $$\eqalign{ & \tan {60^ \circ } = \frac{{DC}}{{BC}} \cr & \sqrt 3 = \frac{{600}}{{BC}} \cr & BC = \frac{{600}}{{\sqrt 3 }}\,..........\,\left( {\text{i}} \right) \cr & \tan {45^ \circ } = \frac{{DC}}{{AC}} \cr & 1 = \frac{{600}}{{AC}} \cr & AC = 600\,............\,\left( {{\text{ii}}} \right) \cr} $$ Distance between the objects $$ = AC = \left( {AC - BC} \right)$$ $$ = 600 - \frac{{600}}{{\sqrt 3 }}$$ [∵ from (1) and (ii)] $$\eqalign{ & = 600\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{600\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{3} \cr & = 200\sqrt 3 \left( {\sqrt 3 - 1} \right) \cr & = 200\left( {3 - \sqrt 3 } \right) \cr & = 200\left( {3 - 1.73} \right) \cr & = 254\,m \cr} $$

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

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 $$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{AC}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{AC}} \cr & \Rightarrow AC = h\sqrt 3 \,.........\,\left( {\text{i}} \right) \cr & \tan {45^ \circ } = \frac{{DC}}{{BC}} \cr & \Rightarrow 1 = \frac{h}{{BC}} \cr & \Rightarrow BC = h\,........\,\left( {{\text{ii}}} \right) \cr & {\text{We know that,}} \cr & AB = \left( {AC - BC} \right) \cr & \Rightarrow 40 = \left( {AC - BC} \right) \cr} $$ $$ \Rightarrow 40 = \left( {h\sqrt 3 - h} \right)$$ [∵ from (1) and (ii)] $$\eqalign{ & \Rightarrow 40 = h\left( {\sqrt 3 - 1} \right) \cr & \Rightarrow h = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \cr & = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{{\left( {3 - 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{2} \cr & = 20\left( {\sqrt 3 + 1} \right) \cr & = 20\left( {1.73 + 1} \right) \cr & = 20 \times 2.73 \cr & = 54.6\,m \cr} $$

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

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 = $$\theta $$ $$\eqalign{ & {\text{From the right}}\Delta PQR, \cr & {\text{tan}}\,\theta = \frac{{RQ}}{{PQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{6\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \cr & \Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {60^ \circ } \cr} $$

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° $$\eqalign{ & \tan {60^ \circ } = \frac{{BA}}{{CA}} \cr & \sqrt 3 = \frac{{BA}}{{150}} \cr & BA = 150\sqrt 3 \cr} $$ i.e, the distance travelled by the balloon = $$150\sqrt 3 $$ meters time taken = 2 min = 2 × 60 = 120 seconds $$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{150\sqrt 3 }}{{120}} = 1.25\sqrt 3 \cr} $$ = 1.25 × 1.73 = 2.16 meter/second

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

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 $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABC, \cr & \tan{22.5^ \circ } = \frac{{AB}}{{BC}}\,.....\left( {eq:1} \right)\, \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,CDE, \cr & \tan {67.5^ \circ } = \frac{{ED}}{{CD}}\,.....\left( {eq:2} \right) \cr & \frac{{\left( {eq:2} \right)}}{{\left( {eq:1} \right)}} \Rightarrow \frac{{\tan {{67.5}^ \circ }}}{{\tan{{22.5}^ \circ }}} = \frac{{\left( {\frac{{ED}}{{CD}}} \right)}}{{\left( {\frac{{AB}}{{BC}}\,} \right)}} \cr & = \frac{{ED}}{{AB}}\,\,\,\,\,\left( {\because CD = BC} \right) \cr & \Rightarrow \frac{{\tan \left( {{{90}^ \circ } - {{22.5}^ \circ }} \right)}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}} \cr & \Rightarrow \frac{{\cot {{22.5}^ \circ }}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}} \cr & \,\,\,\,\,\,\,\,\,\,\left[ {\because \tan \left( {90 - \theta } \right) = \cot \theta } \right] \cr & \Rightarrow \frac{{\left( {\frac{1}{{\tan {{22.5}^ \circ }}}} \right)}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}} \cr & \,\,\,\,\,\,\,\,\,\,\,\left[ {\because \cot \theta = \frac{1}{{\tan \theta }}} \right] \cr & \Rightarrow \frac{{ED}}{{AB}} = \frac{1}{{{{\left( {\tan {{22.5}^ \circ }} \right)}^2}}} \cr & = \frac{1}{{{{\left( {\sqrt 2 - 1} \right)}^2}}} \cr & = {\left( {\frac{1}{{\sqrt 2 - 1}}} \right)^2} \cr & = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}} \right]^2} \cr & = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {2 - 1} \right)}}} \right]^2} \cr & = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{1}} \right]^2} \cr & = {\left( {\sqrt 2 + 1} \right)^2} \cr & = \left( {2 + 2\sqrt 2 + 1} \right) \cr & = \left( {3 + 2\sqrt 2 } \right) \cr & {\text{Required}}\,{\text{Ratio}} \cr & = ED:AB \cr & = \left( {3 + 2\sqrt 2 } \right):1 \cr} $$

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 $$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{BC}} \cr & h = \frac{{BC}}{{\sqrt 3 }}\,......\left( eq : 1 \right) \cr & \tan {60^ \circ } = \frac{{AC}}{{BC}} \cr & \sqrt 3 = \frac{{18 + h}}{{BC}} \cr & 18 + h = BC \times \sqrt 3 \,......\left( eq: 2 \right) \cr & \frac{eq : 1}{eq : 2} \Rightarrow \frac{h}{{18 + h}} = \frac{{\left( {\frac{{BC}}{{\sqrt 3 }}} \right)}}{{\left( {BC \times \sqrt 3 } \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3} \cr & \Rightarrow 3h = 18 + h \cr & \Rightarrow 2h = 18 \cr & \Rightarrow h = 9\,{\text{m}} \cr} $$ 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 $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFD, \cr & \tan {45^ \circ } = \frac{{DF}}{{AF}} \cr & 1 = \frac{{DF}}{5} \cr & DF = 5\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFC, \cr & \tan {60^ \circ } = \frac{{CF}}{{AF}} \cr & \sqrt 3 = \frac{{CF}}{5} \cr & CF = 5\sqrt 3 \,......\,\left( 2 \right) \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{window}} \cr & = CD = \left( {CF - DF} \right) \cr} $$ $$ = 5\sqrt 3 - 5$$ [∵ Substituted the value of CF and DF from (1) and (2)] $$\eqalign{ & = 5\left( {\sqrt 3 - 1} \right) \cr & = 5\left( {1.73 - 1} \right) \cr & = 5 \times 0.73 \cr & = 3.65\,{\text{m}} \cr} $$

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 $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,CDA \cr & \tan {60^ \circ } = \frac{{AD}}{{CD}} \cr & \sqrt 3 = \frac{h}{x}\,......\left( {eq:1} \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,BDA \cr & \tan {30^ \circ } = \frac{{AD}}{{BD}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{70 + x}}\,......\left( {eq:2} \right) \cr & \frac{{eq:1}}{{eq:2}} \Rightarrow \frac{{\sqrt 3 }}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}} = \frac{{\left( {\frac{h}{x}} \right)}}{{\left( {\frac{h}{{70 + x}}} \right)}} \cr & \Rightarrow 3 = \frac{{70 + x}}{x} \cr & \Rightarrow 2x = 70 \cr & \Rightarrow x = 35 \cr} $$ Substituting this value of x in eq : 1, we have $$\eqalign{ & \sqrt 3 = \frac{h}{{35}} \cr & \Rightarrow h = 35\sqrt 3 = 35 \times 1.73 \cr & = 60.55 \approx 60.6 \cr} $$

$$\eqalign{ & \tan {30^ \circ } = \frac{{RQ}}{{PQ}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{200}}{{PQ}} \cr & PQ = 200\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\, = 200 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\, = 346\,{\text{m}} \cr} $$

Let DC be the wall, AB be the tree. Given that ∠DBC = 30°, ∠DAE = 60°, DC = 11 m $$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{11}}{{BC}} \cr & BC = 11\sqrt 3 \,m \cr & AE = BC = 11\sqrt 3 \,m\,.....\left( 1 \right) \cr & \tan {60^ \circ } = \frac{{ED}}{{AE}} \cr} $$ $$\sqrt 3 = \frac{{ED}}{{11\sqrt 3 }}$$ [∵ Substituted value of AE from (1)] $$\eqalign{ & ED = 11\sqrt 3 \times \sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\, = 11 \times 3 \cr & \,\,\,\,\,\,\,\,\,\, = 33 \cr & {\text{Height}}\,{\text{of}}\,{\text{the}}\,{\text{tree}} \cr & = AB = EC = \left( {ED + DC} \right) \cr & = 33 + 11 \cr & = 44\,{\text{m}} \cr} $$

Consider a right-angled triangle PQR as shown below. Let QR = 3 and PR = 5 such that $$\sin \theta = \frac{3}{5}\,\,\,\,\left[ {{\text{i}}{\text{.e}}{\text{.}},\theta = {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right]$$ $$PQ = \sqrt {P{R^2} - Q{R^2}} $$ (∵ Pythagorean theorem) $$\eqalign{ & = \sqrt {{5^2} - {3^2}} \cr & = 4 \cr} $$ $${\text{i}}{\text{.e}}{\text{.}},\,{\text{when}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right),$$ PQ : QR = 4 : 3 ......(eq : 1) Now Let's solve the question. Let P be the point of observation and QR be the tower as shown in the below diagram. $${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$ and PQ = 20 m We know that PQ : QR = 4 : 3 (from eq : 1) $$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}},\,20:QR = 4:3 \cr & \Rightarrow 20 \times 3 = QR \times 4 \cr & \Rightarrow QR = 15\,{\text{m}} \cr} $$ i.e. height of the tower = 15 m Solution 2 Let P be the point of observation and QR be the tower. $${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$ and PQ = 20 m Let the height of the tower, QR = h and PR = x $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta PQR, \cr & \sin \theta = \frac{{QR}}{{PR}} \cr & \Rightarrow \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right] = \frac{h}{x} \cr & \Rightarrow \frac{3}{5} = \frac{h}{x} \cr & \Rightarrow x = \frac{{5h}}{3}\,.....\left( {eq:1} \right) \cr} $$ From Pythagorean theorem, we have $$\eqalign{ & P{Q^2} + Q{R^2} = P{R^2} \cr & {20^2} + {h^2} = {x^2} \cr} $$ $${20^2} + {h^2} = {\left( {\frac{{5h}}{3}} \right)^2}$$ (∵ Substituted the value of x from eq:1) $$\eqalign{ & {20^2} + {h^2} = \frac{{25{h^2}}}{9} \cr & \frac{{16{h^2}}}{9} = {20^2} \cr & \frac{{4h}}{3} = 20 \cr & h = \frac{{3 \times 20}}{4} = 15{\text{m}} \cr} $$ i.e. height of the tower = 15 m

Let BA be the ladder and AC be the wall as shown above. Then the distance of the foot of the ladder from the wall = BC Given that BA = 10 m, ∠ BAC = 60° $$\eqalign{ & \sin {60^ \circ } = \frac{{BC}}{{BA}} \cr & \frac{{\sqrt 3 }}{2} = \frac{{BC}}{{10}} \cr & BC = 10 \times \frac{{\sqrt 3 }}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 5 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\, = 8.65\,{\text{m}} \cr} $$

$$\eqalign{ & \tan {45^ \circ } = \frac{{SR}}{{QR}} \cr & \tan {30^ \circ } = \frac{{SR}}{{PR}} = \frac{{SR}}{{\left( {PQ + QR} \right)}} \cr} $$ Two equations and 3 variables. Hence we can not find the required value with the given data. (Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value.)

Let CB be the pole and point D divides it such that BD : DC = 1 : 9 Given that AB = 15 m Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $$\theta $$ From "Angle Bisector Theorem", we have $$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$ $$ \Rightarrow \frac{1}{9} = \frac{{15}}{{AC}}$$ [∵ BD : DC = 1 : 9 and AB = 15(given)] $$ \Rightarrow AC = 15 \times 9\,{\text{m}}\,......\left( {eq:1} \right)$$ $${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$ $$CB = \sqrt {A{C^2} - A{B^2}} $$ (∵ Pythagorean theorem) $$ = \sqrt {{{\left( {15 \times 9} \right)}^2} - {{15}^2}} $$ [∵AC=15 × 9(eq : 1) and AB=15 m(given)] $$\eqalign{ & = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} \cr & = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} \cr & = \sqrt {{{15}^2} \times 80} \cr & = \sqrt {{{15}^2} \times 16 \times 5} \cr & = 15 \times 4 \times \sqrt 5 \cr & = 60\sqrt 5 \,{\text{m}} \cr} $$

Consider the diagram shown above. Let AB be the tower. Let C and D be the positions of the boat Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m $$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & \Rightarrow 1 = \frac{{AB}}{{100}} \cr & \Rightarrow AB = 100\,......\left( {eq:1} \right) \cr} $$ $$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$ $$ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{BD}}$$ (∵ Substituted the value of AB from equation 1) $$ \Rightarrow BD = 100\sqrt 3 $$ $$\eqalign{ & CD = \left( {BD - BC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {100\sqrt 3 - 100} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 100\left( {\sqrt 3 - 1} \right) \cr} $$ It is given that the distance CD is covered in 10 seconds. i.e., the distance $$100\left( {\sqrt 3 - 1} \right)$$ is covered in 10 seconds. $$\eqalign{ & {\text{Required}}\,{\text{speed}} \cr & = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{10}} \cr & = 10\left( {1.73 - 1} \right) \cr & = 7.3\,{\text{meter/seconds}} \cr & = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = 26.28\,{\text{km/hr}} \cr} $$

Let C and D be the position of the aeroplanes. Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45° $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC, \cr & \tan {60^ \circ } = \frac{{CB}}{{AB}} \cr & \sqrt 3 = \frac{{900}}{{AB}} \cr & AB = \frac{{900}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{900 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{900\sqrt 3 }}{3} \cr & \,\,\,\,\,\,\,\,\,\, = 300\sqrt 3 \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABD, \cr & \tan {45^ \circ } = \frac{{DB}}{{AB}} \cr & 1 = \frac{{DB}}{{AB}} \cr & DB = AB = 300\sqrt 3 \cr & \cr & {\text{Required}}\,{\text{height}} \cr & = CD \cr & = \left( {CB - DB} \right) \cr & = \left( {900 - 300\sqrt 3 } \right) \cr & = \left( {900 - 300 \times 1.73} \right) \cr & = \left( {900 - 519} \right) \cr & = 381\,{\text{m}} \cr} $$

Consider the diagram shown above where QR represents the tree and PQ represents its shadow $$\eqalign{ & {\text{We}}\,{\text{have}},\,QR = PQ \cr & {\text{Let}}\,\angle QPR = \theta \cr} $$ $$\tan \theta = \frac{{QR}}{{PQ}} = 1$$ (since QR = PQ) $$ \Rightarrow \theta = {45^ \circ }$$ i.e., required angle of elevation = 45°

Let BD be the tower and A and C be the positions of the persons. Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60° $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABD, \cr & \tan {30^ \circ } = \frac{{BD}}{{BA}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{50}}{{BA}} \cr & \Rightarrow BA = 50\sqrt 3 \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CBD, \cr & \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{{50}}{{BC}} \cr & \Rightarrow BC = \frac{{50}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 }}{3} \cr} $$ Distance between the two persons $$\eqalign{ & = AC = BA + BC \cr & = 50\sqrt 3 + \frac{{50\sqrt 3 }}{3} \cr & = \sqrt 3 \left( {50 + \frac{{50}}{3}} \right) \cr & = \frac{{200\sqrt 3 }}{3}\,{\text{m}} \cr} $$ i.e., the distance travelled by the car in 10 seconds = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$ $$\eqalign{ & {\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}} \cr & = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = {\text{24}}\sqrt 3 \,{\text{km/hr}} \cr} $$

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car. Then, ∠ ADC = 30° , ∠ ACB = 45° Let AB = h, BC = x, CD = y $$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{x} \cr & \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow h = x\,......\left( 1 \right) \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AB}}{{\left( {BC + CD} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{x + y}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{x + y}} \cr & \Rightarrow x + y = \sqrt 3 \,h \cr & \Rightarrow y = \sqrt 3 \,h - x \cr} $$ $$ \Rightarrow y = \sqrt 3 \,h - h$$ (∵ Substituted the value of x from equation 1 ) $$ \Rightarrow y = h\left( {\sqrt 3 - 1} \right)$$ Given that distance y is covered in 8 minutes. i.e, distance $$h\left( {\sqrt 3 - 1} \right)$$ is covered in 8 minutes. Time to travel distance x = Time to travel distance h (∵ Since x = h as per equation 1). Let distance h is covered in t minutes. since distance is proportional to the time when the speed is constant, we have $$\eqalign{ & h\left( {\sqrt 3 - 1} \right) \propto 8\,......\left( {\text{A}} \right) \cr & h \propto t\,..............\left( {\text{B}} \right) \cr & \frac{{\left( {\text{A}} \right)}}{{\left( {\text{B}} \right)}} \Rightarrow \frac{{h\left( {\sqrt 3 - 1} \right)}}{h} = \frac{8}{t} \cr & \Rightarrow \left( {\sqrt 3 - 1} \right) = \frac{8}{t} \cr & \Rightarrow t = \frac{8}{{\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{\left( {1.73 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{.73}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{800}}{{73}}\,{\text{minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 10\frac{{70}}{{73}}\,{\text{minutes}} \cr & \approx {\text{10}}\,{\text{minutes}}\,{\text{57}}\,{\text{seconds}} \cr} $$

Let AB and CD be the poles with heights h and 2h respectively. Given that distance between the poles, BD = 200 m Let E be the middle point of BD, ∠ AEB = $$\theta $$ ∠ CED = (90 - $$\theta $$) (∵ given that angular elevations are complementary) Since E is the middle point of BD, we have BE = ED = 100 m $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABE, \cr & \tan \theta = \frac{{AB}}{{BE}} \cr & \tan \theta = \frac{h}{{100}} \cr & h = 100\tan \theta \,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta EDC, \cr & \tan\left( {90 - \theta } \right) = \frac{{CD}}{{ED}} \cr} $$ $$\cot \theta = \frac{{2h}}{{100}}$$ $$\left[ {\because \tan \left( {90 - \theta } \right) = \cot \theta } \right]$$ $$\eqalign{ & 2h = 100\cot \theta \,......\left( 2 \right) \cr & \left( 1 \right) \times \left( 2 \right) \cr} $$ $$ \Rightarrow 2{h^2} = {100^2}$$ $$\left[ {\because \tan \theta \times \cot\theta = \tan\theta \times \frac{1}{{\tan \theta }} = 1} \right]$$ $$\eqalign{ & \Rightarrow \sqrt 2 \,h = 100 \cr & \Rightarrow h = \frac{{100}}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 50\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 50 \times 1.41 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 70.5 \cr & \Rightarrow 2h = 2 \times 70.5 = 141 \cr} $$ i.e. the height of the poles are 70.5 m and 141 m.

Let ED be the building and AC be the tower. Given that ED = 230 m, ∠ADC = b, ∠AEB = a Also given that tan a = $$\frac{5}{{12}}$$ and tan b = $$\frac{4}{{5}}$$ Let AC = h Required Distance = Distance between the top of these buildings = AE $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABE, \cr & \tan \left( a \right) = \frac{{AB}}{{BE}} \cr} $$ $$ \Rightarrow \frac{5}{{12}} = \frac{{\left( {h - 230} \right)}}{{BE}}$$ [∵ tan(a) = $$\frac{5}{{12}}$$ (given), AB = (AC - BC) = (AC - ED) = (h - 230)] $$\eqalign{ & \Rightarrow BE = \frac{{12\left( {h - 230} \right)}}{5}\,.....\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ACD, \cr & \tan \left( b \right) = \frac{{AC}}{{CD}} \cr} $$ $$ \Rightarrow \frac{4}{5} = \frac{h}{{CD}}$$ [∵ tan(b) = $$\frac{4}{5}$$ (given), AC = h] $$\eqalign{ & \Rightarrow CD = \frac{{5h}}{4}\,.....\left( 2 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{diagram,}}\,BE = CD \cr} $$ $$ \Rightarrow \frac{{12\left( {h - 230} \right)}}{5} = \frac{{5h}}{4}$$ (from equation 1 & 2) $$\eqalign{ & \Rightarrow 48h - \left( {4 \times 12 \times 230} \right) = 25h \cr & \Rightarrow 23h = \left( {4 \times 12 \times 230} \right) \cr & \Rightarrow h = \frac{{\left( {4 \times 12 \times 230} \right)}}{{23}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 480\,{\text{m}}\,......\left( 3 \right) \cr & AB = \left( {AC - BC} \right) \cr} $$ $$\,\,\,\,\,\,\,\,\,\,\,\, = \left( {480 - 230} \right)$$ [∵ Since AC = h = 480 (from 3) and BC = ED = 230 m(given)] $$\,\,\,\,\,\,\,\,\,\,\,\, = 250\,{\text{m}}$$ In the triangle ABE, tan(a) =$$\frac{5}{{12}}$$ . Let's figure out the value of sin(a) now. Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = $$\frac{5}{{12}}$$ $$\eqalign{ & {\text{hypotenuse}} = \sqrt {{5^2} + {{12}^2}} = 13 \cr & {\text{i}}{\text{.e}}{\text{.,}} \cr & \sin \left( a \right) = \frac{{{\text{opposite}}\,{\text{side}}}}{{{\text{hypotenuse}}}} = \frac{5}{{13}} \cr} $$ We have seen that $$\sin \left( a \right) = \frac{5}{{13}}$$ $$\eqalign{ & \Rightarrow \frac{{AB}}{{AE}} = \frac{5}{{13}} \cr & \Rightarrow AE = AB \times \frac{{13}}{5} \cr & = 250 \times \frac{{13}}{5} \cr & = 650\,{\text{m}} \cr} $$ i.e., Distance between the top of the buildings = 650 m

Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower. Given that height of the tower, AC = 80 m and the angle of depression, ∠DAB = 30° ∠ABC = ∠DAB = 30° (because DA || BC) $$\eqalign{ & \tan {30^ \circ } = \frac{{AC}}{{BC}} \cr & \Rightarrow \tan {30^ \circ } = \frac{{80}}{{BC}} \cr & \Rightarrow BC = \frac{{80}}{{\tan {{30}^ \circ }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{80}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}} \cr & = 80 \times 1.73 = 138.4\,{\text{m}} \cr} $$ i.e., Distance of the bus from the foot of the tower = 138.4 m

Let A be the foot and C be the summit of a mountain. Given that ∠CAB = 45° From the diagram, CB is the height of the mountain. Let CB = x Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that ∠DAY = 30° It is also given that from the point D, the elevation is 60° i.e., ∠CDE = 60° $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC, \cr & \tan {45^ \circ } = \frac{{CB}}{{AB}} \cr} $$ $$ \Rightarrow 1 = \frac{x}{{AB}}$$ [∵ CB = x(the height of the mountain)] $$\eqalign{ & \Rightarrow AB = x\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta AYD, \cr & \sin {30^ \circ } = \frac{{DY}}{{AD}} \cr} $$ $$ \Rightarrow \frac{1}{2} = \frac{{DY}}{2}$$ (∵ Given that AD = 2) $$\eqalign{ & \Rightarrow DY = 1\,.......\left( 2 \right) \cr & \cos {30^ \circ } = \frac{{AY}}{{AD}} \cr} $$ $$ \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{AY}}{2}$$ (∵ Given that AD = 2) $$\eqalign{ & \Rightarrow AY = \sqrt 3 \,......\left( 3 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CED, \cr & \tan {60^ \circ } = \frac{{CE}}{{DE}} \cr} $$ $$ \Rightarrow \tan {60^ \circ } = \frac{{\left( {CB - EB} \right)}}{{YB}}$$ [∵ CE = (CB - EB) and DE = YB)] $$ \Rightarrow \tan {60^ \circ } = \frac{{\left( {CB - DY} \right)}}{{AB - AY}}$$ [∵ EB = DY and YB = (AB - AY)] $$ \Rightarrow \tan {60^ \circ } = \frac{{\left( {x - 1} \right)}}{{\left( {x - \sqrt 3 } \right)}}$$ [∵ CB = x, DY = 1(eq:2), AB = x(eq:1) and AY = $${\sqrt 3 }$$ (eq:3)] $$\eqalign{ & \Rightarrow \sqrt 3 = \frac{{\left( {x - 1} \right)}}{{\left( {x - \sqrt 3 } \right)}} \cr & \Rightarrow x\sqrt 3 - 3 = x - 1 \cr & \Rightarrow x\left( {\sqrt 3 - 1} \right) = 2 \cr & \Rightarrow 0.73x = 2 \cr & \Rightarrow x = \frac{2}{{0.73}} = 2.7 \cr} $$ i.e., the height of the mountain = 2.7 km

Let AD be the flagstaff and CD be the building. Assume that the flagstaff and building subtend equal angles at point B. Given that AD = 50 m, CD = h and BC = 200 m Let ∠ABD = $$\theta $$, ∠DBC = $$\theta $$ (∵ flagstaff and building subtend equal angles at a point on level ground). Then, ∠ABC = 2$$\theta $$ $$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta BCD, \cr & \tan \theta = \frac{{DC}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{200}}\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta BCA, \cr & \tan 2\theta = \frac{{AC}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AD + DC}}{{200}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + h}}{{200}} \cr} $$ $$ \Rightarrow \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{50 + h}}{{200}}$$ $$\left( {\because \tan \left( {2\theta } \right) = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$$ $$\frac{{2\left( {\frac{h}{{200}}} \right)}}{{1 - \frac{{{h^2}}}{{{{200}^2}}}}} = \frac{{50 + h}}{{200}}$$ (∵ substituted value of tan $$\theta $$ from eq:1) $$\eqalign{ & \Rightarrow 2h = \left( {1 - \frac{{{h^2}}}{{{{200}^2}}}} \right)\,\left( {50 + h} \right) \cr & \Rightarrow 2h = 50 + h - \frac{{50{h^2}}}{{{{200}^2}}} - \frac{{{h^3}}}{{{{200}^2}}} \cr} $$ $$ \Rightarrow 2\left( {{{200}^2}} \right)h = 50{\left( {200} \right)^2} + $$ $$h{\left( {200} \right)^2} - $$ $$50{h^2} - {h^3}$$ (∵ multiplied LHS and RHS by $${{{200}^2}}$$ ) h 3 + 50h 2 + (200) 2 h - (200) 2 50 = 0

Consider the diagram shown above. AC represents the hill and DE represents the pole Given that AC = 100 m ∠XAD = ∠ADB = 30° (∵ AX || BD ) ∠XAE = ∠AEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100 - h) (∵ AC = 100 and BC = h), BD = CE $$\eqalign{ & \tan {60^ \circ } = \frac{{AC}}{{CE}} \cr & \Rightarrow \sqrt 3 = \frac{{100}}{{CE}} \cr & \Rightarrow CE = \frac{{100}}{{\sqrt 3 }}\,......\left( 1 \right) \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100 - h}}{{BD}} \cr} $$ $$ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100 - h}}{{\left( {\frac{{100}}{{\sqrt 3 }}} \right)}}$$ (∵ BD = CE and substituted the value of CE from eq. 1) $$\eqalign{ & \Rightarrow \left( {100 - h} \right) = \frac{1}{{\sqrt 3 }} \times \frac{{100}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 33.33 \cr & \Rightarrow h = 100 - 33.33 = 66.67\,{\text{m}} \cr} $$ i.e., the height of the pole = 66.67 m

$$\eqalign{ & {\text{in}}\,\Delta PNQ,\,\tan {45^ \circ } = \frac{{PQ}}{{NQ}} \cr & \therefore PQ = NQ = 50{\text{m}} \cr & {\text{in}}\,\Delta MNQ,\,\tan {30^ \circ } = \frac{{MQ}}{{NQ}} \cr & \therefore \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{50}} \cr & \therefore MQ = \frac{{50}}{{\sqrt 3 }} \cr & \therefore h = PM \cr & \,\,\,\,\,\,\,\,\,\, = PQ - MQ \cr & \,\,\,\,\,\,\,\,\,\, = 50 - \frac{{50}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 - 1} \right) \cr} $$ ∴ Poster height = $$\frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 - 1} \right)$$

$$\eqalign{ & {\text{Height}}\,{\text{of}}\,{\text{pole}} = PQ \cr & \tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{m} \cr & \tan {45^ \circ } = 1 = \frac{{PQ}}{n} \cr & {\text{Multiply}}\,{\text{both}}\,{\text{equations}} \cr & \sqrt 3 \times 1 = \frac{{PQ}}{m} \times \frac{{PQ}}{n} \cr & \therefore PQ = \sqrt {mn\sqrt 3 } \,{\text{units}} \cr} $$

$$\eqalign{ & {\text{in}}\,\Delta MNQ,\tan {30^ \circ } = \frac{{MQ}}{{NQ}} \cr & \therefore \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{10}} \cr & \therefore MQ = \frac{{10}}{{\sqrt 3 }} \cr & {\text{Also}}\,{\text{by}}\,{\text{Pythagoras}}\,{\text{theorem}} \cr & M{N^2} = M{Q^2} + N{Q^2} \cr & \therefore {L^2} = \frac{{100}}{3} + 100 \cr & \therefore L = \frac{{20}}{{\sqrt 3 }} \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tree}} \cr & = L + MQ \cr & = \frac{{20}}{{\sqrt 3 }} + \frac{{10}}{{\sqrt 3 }} \cr & = \frac{{30}}{{\sqrt 3 }} \cr & = \frac{{3 \times 10}}{{\sqrt 3 }} \cr & = 10\sqrt 3 \,{\text{m}} \cr} $$

$$\eqalign{ & {\text{Original tree height}} = {\text{h}} = {\text{MQ}} \cr & {\text{New}}\,{\text{tree}}\,{\text{height}} = PQ \cr & {\text{in}}\,\Delta MQN,\,\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{NQ}} \cr & \therefore MQ = \frac{{300}}{{\sqrt 3 }} \cr & {\text{in}}\,\Delta PQN,\,\tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{NQ}} \cr & \therefore PQ = 300\sqrt 3 \cr & {\text{Tree}}\,{\text{grew}} = PQ - MQ \cr & \therefore {\text{Tree}}\,{\text{grew}} \cr & = 300\sqrt 3 - \frac{{300}}{{\sqrt 3 }} \cr & = 300\frac{2}{{\sqrt 3 }} \cr & = 3 \times 100 \times \frac{2}{{\sqrt 3 }} \cr & = 200\sqrt 3 \,{\text{m}} \cr} $$

Let PQ be three and M and N be positions where Mphan stands. $$\eqalign{ & {\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{MQ}} \cr & \therefore PQ = MQ \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{NQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{10 + MQ}} \cr & \therefore 10 + MQ = \sqrt 3 \,PQ \cr} $$ $$\therefore 10 + PQ = \sqrt 3 \,PQ$$ (As, PQ = MQ) $$\therefore PQ = \frac{{10}}{{\sqrt 3 - 1}}\,{\text{cm}}$$

$$\eqalign{ & {\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{NQ}} \cr & \therefore {\text{Tree}}\,{\text{height}} \cr & = PQ = NQ \cr & = \left( {240 + MQ} \right) \cr & \tan {60^ \circ } = \sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{MQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 + MQ}}{{MQ}} \cr & \therefore 240 + MQ = \sqrt 3 \,MQ \cr & \therefore MQ = \frac{{240}}{{\sqrt 3 - 1}}\,{\text{ft}} \cr & \therefore NQ = 240 + MQ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240\sqrt 3 }}{{\sqrt 3 - 1}}\,{\text{ft}} \cr} $$ = Rohit was this much far away initially

Let the tree break at height h cm from ground at point M. The broken part makes angle of 30° ∴ Broken Part MP = MN = 24 - h $$\eqalign{ & {\text{in}}\,\Delta MCQ, \cr & \sin {30^ \circ } = \frac{1}{2} = \frac{{MQ}}{{MN}} = \frac{{\text{h}}}{{24 - {\text{h}}}} \cr & \therefore 24 - {\text{h}} = 2{\text{h}} \cr} $$ ∴ h = 8 cm = Tree breaks at this height

In ∠ABR, ∠ARB = 90° and ∠BAR = 45° Sum of angles of a triangle = 180° So ∠ABR = 180 - 90 - 45 = 45° ∴ BR = AR AS = RQ = 10m Also, BR = BQ - RQ = 25 - 10 = 15m ∴ AR = 15m = Distance between houses