Practice MCQ Questions and Answer on Height and Distance

Solution: $$\eqalign{ & 3 \to 240 \cr & \sqrt 3 \to \frac{{240}}{3} \times \sqrt 3 = 80\sqrt 3 \cr & {\text{Height of the pole}} \cr & = 160 - 80\sqrt 3 \cr & = 80\left( {2 - \sqrt 3 } \right) \cr} $$

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: $$\eqalign{ & \sqrt 3 \to 120\sqrt 3 \cr & 1{\text{u}} \to 120 \cr & 2{\text{u}} \to 120 \times 2 = 240{\text{m}} \cr} $$

Solution: \[\begin{gathered} \begin{array}{*{20}{c}} {\sec \theta = }&{\frac{{\,\,\,13}}{\begin{gathered} \,\,\,12\,\, \hfill \\ ^{ \times 3} \downarrow \hfill \\ \end{gathered} }} \\ {}&{\,\,\,36} \end{array} \hfill \\ 5{\text{ unit}} \to 5 \times 3 = 15{\text{ m}} \hfill \\ \end{gathered} \]

Solution: 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 = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$ $$\eqalign{ & {\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}} \cr & = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = {\text{24}}\sqrt 3 \,{\text{km/hr}} \cr} $$

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

Solution: Angle of Depression = Angle of Elevation Tower PS = 10 m in height $$\eqalign{ & {\text{tan}}{45^ \circ } = 1 = \frac{{PS}}{{RS}} \cr & \therefore PS = RS = 10 \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PS}}{{SQ}} = \frac{{10}}{{SQ}} \cr & \therefore SQ = 10\sqrt 3 \cr & RQ = RS + SQ \cr & \,\,\,\,\,\,\,\,\,\,\, = 10 + 10\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\,\, = 27.32\,{\text{m}} \cr} $$

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

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