Practice MCQ Questions and Answer on Height and Distance

Solution: 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 $$\eqalign{ & \sqrt 3 = \frac{h}{{35}} \cr & \Rightarrow h = 35\sqrt 3 = 35 \times 1.73 \cr & = 60.55 \approx 60.6 \cr} $$

Solution: $$\eqalign{ & AB = 10\tan {60^ \circ } = 10\sqrt 3 \cr & BC = 10\sec {60^ \circ } = 20 \cr} $$ Since, length of ladder will remain same, hence BC = DE = 20 $$\eqalign{ & {\text{In }}\Delta ADE,\,DE = 20{\text{ ft}} \cr & \Rightarrow \frac{{AD}}{{DE}} = \cos {30^ \circ } \cr & \Rightarrow AD = 20 \times \frac{{\sqrt 3 }}{2} = 10\sqrt 3 \cr & \Rightarrow x = AD - AC \cr & \Rightarrow x = 10\sqrt 3 - 10 \cr & \Rightarrow x = 10\left( {\sqrt 3 - 1} \right){\text{ft}} \cr} $$

Solution: 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)] $$\eqalign{ & \Rightarrow 40 = h\left( {\sqrt 3 - 1} \right) \cr & \Rightarrow h = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \cr & = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{{\left( {3 - 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{2} \cr & = 20\left( {\sqrt 3 + 1} \right) \cr & = 20\left( {1.73 + 1} \right) \cr & = 20 \times 2.73 \cr & = 54.6\,m \cr} $$

Solution: $$\eqalign{ & {\text{In}}\,\Delta PBQ,\,\tan {60^ \circ } = \frac{{PQ}}{{BQ}} \cr & \therefore BQ = \frac{{PQ}}{{\sqrt 3 }} \cr & {\text{In}}\,\Delta PAQ,\,\tan {30^ \circ } = \frac{{PQ}}{{AQ}} \cr & \therefore \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{50 + BQ}} \cr & \therefore PQ = \frac{{50 + BQ}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + \frac{{PQ}}{{\sqrt 3 }}}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 + PQ}}{{\sqrt 3 \times \sqrt 3 }} \cr & \therefore 3PQ = 50\sqrt 3 + PQ \cr} $$ $$\therefore PQ = 25\sqrt 3 \,{\text{m}} = $$     Height of tree

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 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: Length of the ladder = 6.6√2 m

Solution: $$\eqalign{ & {\text{In }}\Delta ABD, \cr & \tan {30^ \circ } = \frac{P}{B} \cr & \frac{1}{{\sqrt 3 }} = \frac{{24}}{{BD}} \cr & BD = 24\sqrt 3 \cr & BD = AE = 24\sqrt 3 \cr & {\text{In }}\Delta AEC, \cr & \tan {60^ \circ } = \frac{P}{B} \cr & \sqrt 3 = \frac{{CE}}{{AE}} \cr & \sqrt 3 = \frac{{CE}}{{24\sqrt 3 }} \cr & CE = 24 \times 3 = 72 \cr & {\text{Height of tower }}CD = CE + ED \cr & = 72 + 24 = 96{\text{ cm}} \cr} $$

Solution: 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 $$\eqalign{ & {\text{Let}}\,AB = h \cr & CE \,\, || \,\, DB \cr & \therefore EC = DB = x{\text{ }}\left( {{\text{suppose}}} \right) \cr & EB = CD = 25 \cr & \therefore AE = h - 25 \cr & {\text{Now in right }}\Delta CDB, \cr & \tan \theta = \frac{{CD}}{{DB}} = \frac{{25}}{x}\,......\left( {\text{i}} \right) \cr & {\text{and in right }}\Delta CAE \cr & \tan \theta = \frac{{AE}}{{CE}} = \frac{{h - 25}}{x}\,......\left( {{\text{ii}}} \right) \cr & {\text{From}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{25}}{x} = \frac{{h - 25}}{x} \cr & \Rightarrow 25 = h - 25 \cr & \Rightarrow h = 25 + 25 = 50 \cr & \therefore {\text{Height of tower}} = 50\,m \cr} $$

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