Practice MCQ Questions and Answer on Height and Distance

Solution: In ΔABC, By pythagoras theorem- (BC)2 = (23)2 - ($$3\sqrt 5 $$ )2 (BC)2 = 529 - 45 (BC)2 = 484 BC = 22 Again ΔCDE, (CD)2 = (23)2 - ($${\text{4}}\sqrt {15} $$ )2 (CD)2 = 529 - 240 (CD)2 = 289 CD = 17 Then width of the street BD = BC + CD = 22 + 17 = 39

Solution: Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60° Let height of the tower = h $$\eqalign{ & {\text{Then in right }}\Delta ABP \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{100}} \Rightarrow \sqrt 3 = \frac{h}{{100}} \cr & \Rightarrow h = 100\sqrt 3 \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = 100\sqrt 3 \cr} $$

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: 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

Solution: 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

Solution: 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

Solution: 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. $$\eqalign{ & {\text{Required}}\,{\text{speed}} \cr & = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{10}} \cr & = 10\left( {1.73 - 1} \right) \cr & = 7.3\,{\text{meter/seconds}} \cr & = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = 26.28\,{\text{km/hr}} \cr} $$

Solution: Length of the ladder = 6.6√2 m

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