Practice MCQ Questions and Answer on Height and Distance

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

Solution: $$\eqalign{ & {\text{In }}\Delta ABC \cr & \tan \theta = \frac{{AB}}{{BC}} \cr & \tan \theta = \frac{{57}}{{19\sqrt 3 }} \cr & \tan \theta = \frac{3}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \therefore \boxed{\theta = {{60}^ \circ }} \cr} $$

Solution: $$\eqalign{ & AB = {\text{hill}} = 200{\text{ metre}} \cr & CD = {\text{tower}} \cr & {\text{In }}\Delta \,APC \cr & \tan {30^ \circ } = \frac{{AP}}{{PC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{AP}}{{PC}} \cr & \Rightarrow AP:PC = 1:\sqrt 3 \,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{In }}\Delta \,ABD \cr & \tan {60^ \circ } = \frac{{AB}}{{BD}} \cr & \sqrt 3 = \frac{{AB}}{{BD}} \cr & \Rightarrow AB:BD = \sqrt 3 :1\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & PB = CD{\text{ and }}PC = BD \cr & {\text{Now}} \cr} $$ \[\begin{array}{*{20}{c}} {AB}&:&{BD}&:&{AP} \\ {\sqrt 3 }&:&1&{}&{} \\ {}&{}&{\sqrt 3 }&:&1 \\ 3&:&{\sqrt 3 }&:&1 \end{array}\] $$\eqalign{ & CD = PB \Rightarrow AB - AP \cr & CD = 3 - 1 \cr & CD = 2{\text{ units}} \cr & AB = 3{\text{ units}} = 200{\text{ metre}} \cr & CD = 2{\text{ units}} = \frac{{200}}{3} \times 2 = 133\frac{1}{3}\,{\text{metre}} \cr} $$

Solution: $$\eqalign{ & \tan {45^ \circ } = \frac{{SR}}{{QR}} \cr & \tan {30^ \circ } = \frac{{SR}}{{PR}} = \frac{{SR}}{{\left( {PQ + QR} \right)}} \cr} $$ 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.)

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: Let AB be the building and EC be the tower. $$\eqalign{ & {\text{In }}\Delta ADE \cr & \tan {60^ \circ } = \frac{{ED}}{{AD}} \cr & \sqrt 3 = \frac{{ED}}{{16}} \cr & ED = 16\sqrt 3 = 16 \times 1.73 = 27.68\,{\text{m}} \cr} $$ Hence the height of the tower EC = CD + ED = 12 + 27.68 = 39.68 m

Solution: 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)] $$\eqalign{ & = 600\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{600\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{3} \cr & = 200\sqrt 3 \left( {\sqrt 3 - 1} \right) \cr & = 200\left( {3 - \sqrt 3 } \right) \cr & = 200\left( {3 - 1.73} \right) \cr & = 254\,m \cr} $$

Solution: P : Q 1 : √3

Solution: 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

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