Practice MCQ Questions and Answer on Height and Distance

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

Solution: $$\eqalign{ & \Delta ABE \cr & \tan {30^ \circ } = \frac{{AB}}{{EB}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 \times h \cr & \Delta EBD \cr & \tan {60^ \circ } = \frac{{BD}}{{EB}} \cr & \sqrt 3 = \frac{{h + 30 + 30}}{x} \cr & \sqrt 3 = \frac{{h + 60}}{{\sqrt 3 h}} \cr & 3h = h + 60 \cr & 2h = 60 \cr & h = 30{\text{m}} \cr & {\text{Height from water surface}} \cr & = 30 + 30 \cr & = 60{\text{m}} \cr} $$

Solution: Length of the ladder = 6.6√2 m

Solution: $$\eqalign{ & \tan {30^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{BD}}{{50}} \cr & \Rightarrow BD = \frac{{50}}{{\sqrt 3 }} \cr & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \cr & \Rightarrow AB = 50 \cr & \therefore AD = AB - BD \cr & = 50 - \frac{{50}}{{\sqrt 3 }} \cr & = \frac{{50\left( {\sqrt 3 - 1} \right)\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & = \frac{{50}}{3}\left( {3 - \sqrt 3 } \right){\text{m}} \cr & 200\% {\text{ of }}AD = \frac{{50}}{3}\left( {3 - \sqrt 3 } \right) \times \frac{{200}}{{100}} \cr & = \frac{{100}}{3}\left( {3 - \sqrt 3 } \right){\text{m}} \cr} $$

Solution: 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

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

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

Solution: 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) $$\therefore PQ = \frac{{10}}{{\sqrt 3 - 1}}\,{\text{cm}}$$

Solution: $$\eqalign{ & {\text{In }}\Delta ABM, \cr & \tan \left( {90 - \theta } \right) = \frac{P}{B} = \frac{{AB}}{{72}} \cr & \frac{1}{{\cot \theta }} = \frac{{72}}{{AB}} \cr & \tan \theta = \frac{{72}}{{AB}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {\text{In }}\Delta ABN, \cr & \tan \theta = \frac{P}{B} = \frac{{AB}}{{128}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{72}}{{AB}} = \frac{{AB}}{{128}} \cr & A{B^2} = 128 \times 72 \cr & AB = \sqrt {16 \times 4 \times 2 \times 36 \times 2} \cr & AB = 4 \times 4 \times 6 \cr & AB = 96 \cr} $$

Solution: 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 $${{{200}^2}}$$ ) h3 + 50h2 + (200)2h - (200)250 = 0