JEE Mathematics Heights and Distances MCQs Set 02

Answer: (b) 16

Explanation:
1. With heights \( p=20 \), \( q=80 \) and base distance \( d \), the point of intersection has height \( h=\dfrac{pq}{p+q} \) (independent of \( d \)).
2. \( h=\dfrac{20\cdot80}{100}=16 \) m. Answer: (b)

Teacher's Note:
1. Similar triangles give \( \dfrac1h=\dfrac1p+\dfrac1q \).

Answer: (d) \( 5(2+\sqrt3) \)

Explanation:
1. The difference in heights is \( 5 \) m, so \( \tan15^\circ=\dfrac5d \).
2. \( d=\dfrac5{2-\sqrt3}=5(2+\sqrt3) \) m. Answer: (d)

Teacher's Note:
1. \( \tan15^\circ=2-\sqrt3 \).

Answer: (b) 20

Explanation:
1. Let \( P \) be the mid-point of \( BC \), \( h \) the height and \( l=AP \). Then \( \cot\beta=\dfrac lh=3\sqrt2 \) gives \( l=3\sqrt2h \).
2. \( \mathrm{cosec}\,\alpha=2\sqrt2 \) gives \( \cot^2\alpha=7 \) and \( BP=\sqrt{100^2-l^2}=h\cot\alpha \), so \( 100^2-18h^2=7h^2 \) and \( h=20 \) m. Answer: (b)

Teacher's Note:
1. \( AP\perp BC \) in an isosceles triangle.

Answer: (b) \( \dfrac{25}{\sqrt3} \)

Explanation:
1. \( x^2+y^2=4 \) gives \( x\dfrac{dx}{dt}=-y\dfrac{dy}{dt} \). At \( y=1 \), \( x=\sqrt3 \).
2. \( \dfrac{dx}{dt}=-\dfrac1{\sqrt3}(-25)=\dfrac{25}{\sqrt3} \) cm/s. Answer: (b)

Teacher's Note:
1. Differentiate the Pythagoras relation with respect to time.

Answer: (a) \( 15(3+\sqrt3) \)

Explanation:
1. Let the tower height be \( h \) and distance \( y \). From \( A \): \( h=y \). From \( B \): \( h-30=\dfrac y{\sqrt3} \).
2. \( y\left(1-\dfrac1{\sqrt3}\right)=30 \), so \( y=\dfrac{30\sqrt3}{\sqrt3-1}=15\sqrt3(\sqrt3+1)=15(3+\sqrt3) \). Answer: (a)

Teacher's Note:
1. Two right triangles share the horizontal distance.

Answer: (a) \( \dfrac23\sqrt{21} \)

Explanation:
1. By Apollonius, \( BD^2=\dfrac{2\cdot5^2+2\cdot7^2-6^2}4=28 \), so \( BD=2\sqrt7 \).
2. Height \( =BD\tan30^\circ=\dfrac{2\sqrt7}{\sqrt3}=\dfrac{2\sqrt{21}}3 \) m. Answer: (a)

Teacher's Note:
1. The median length comes from Apollonius' theorem.

Answer: (a) 50

Explanation:
1. Let the cloud be \( x \) m above \( P \) at horizontal distance \( y \). Then \( \tan30^\circ=\dfrac xy \) and \( \tan60^\circ=\dfrac{25+(25+x)}y \).
2. \( y=\sqrt3x \), so \( \sqrt3=\dfrac{50+x}{\sqrt3x} \), giving \( 3x=50+x \), \( x=25 \). Height above the lake \( =25+25=50 \) m. Answer: (a)

Teacher's Note:
1. The reflection is as far below the surface as the cloud is above.

Answer: (a) 100

Explanation:
1. Let the tower be \( TM \) with height \( h \). Then \( PM=h \) (elevation \( 45^\circ \)) and \( QM=\sqrt3h \).
2. \( PM\perp QR \), so \( h^2+3h^2=200^2 \), giving \( h=100 \) m. Answer: (a)

Teacher's Note:
1. \( PM \) is the altitude of the isosceles triangle.

Answer: (c) \( \dfrac29 \)

Explanation:
1. Let \( AB=h \), \( AP=2h \), \( \angle CPA=\alpha \). Then \( \tan(\alpha+\beta)=\dfrac h{2h}=\dfrac12 \) and \( \tan\alpha=\dfrac{h/2}{2h}=\dfrac14 \).
2. \( \tan\beta=\dfrac{\frac12-\frac14}{1+\frac18}=\dfrac{1/4}{9/8}=\dfrac29 \). Answer: (c)

Teacher's Note:
1. \( \beta=(\alpha+\beta)-\alpha \).

Answer: (d) 5

Explanation:
1. Let \( AB=x \), \( BC=y \) (\( C \) the foot), height \( h \). \( h=(x+y)\tan30^\circ \) and \( h=y\tan60^\circ \), so \( x+y=3y \) and \( y=\dfrac x2 \).
2. \( x \) takes 10 min, so \( y \) takes 5 min. Answer: (d)

Teacher's Note:
1. Speed is uniform, so time is proportional to distance.

Answer: (a) \( \sqrt3:1 \)

Explanation:
1. With height \( h \) and foot \( D \): \( AD=\sqrt3h \), \( BD=h \), \( CD=\dfrac h{\sqrt3} \).
2. \( AB:BC=(\sqrt3h-h):\left(h-\dfrac h{\sqrt3}\right)=(\sqrt3-1):\dfrac{\sqrt3-1}{\sqrt3}=\sqrt3:1 \). Answer: (a)

Teacher's Note:
1. Express each distance via the height.

Answer: (d) \( 20(\sqrt3-1) \)

Explanation:
1. Initially \( OB_1=20\cot45^\circ=20 \). Later \( OB_2=20\cot30^\circ=20\sqrt3 \).
2. Distance flown \( =20\sqrt3-20 \) in 1 s, so speed \( =20(\sqrt3-1) \) m/s. Answer: (d)

Teacher's Note:
1. The height stays at 20 m.

Answer: (c) \( \dfrac{7\sqrt3}2(\sqrt3+1) \) m

Explanation:
1. \( BC=h\cot60^\circ \) and \( BD=h\cot45^\circ=h \), so \( h-\dfrac h{\sqrt3}=7 \).
2. \( h=\dfrac{7\sqrt3}{\sqrt3-1}=\dfrac{7\sqrt3(\sqrt3+1)}2 \) m. Answer: (c)

Teacher's Note:
1. \( BD-BC=CD=7 \).

Answer: (c) \( \dfrac a{\sqrt3} \)

Explanation:
1. \( OA=OB \) and \( \angle AOB=60^\circ \), so \( \triangle OAB \) is equilateral and \( OA=a \).
2. \( \tan30^\circ=\dfrac ha \), so \( h=\dfrac a{\sqrt3} \). Answer: (c)

Teacher's Note:
1. The radius equals the chord \( AB \).

Answer: (a) 20 m

Explanation:
1. Let the breadth be \( x \). Height \( h=x\tan60^\circ=\sqrt3x \) and \( h=(x+40)\tan30^\circ \).
2. \( \sqrt3x=\dfrac{x+40}{\sqrt3} \) gives \( 3x=x+40 \), so \( x=20 \) m. Answer: (a)

Teacher's Note:
1. Equate the two expressions for the height.

Answer: (b) 40 m

Explanation:
1. Let the height be \( h \). The lower quarter subtends \( \theta_1 \) with \( \tan\theta_1=\dfrac{h/4}{40}=\dfrac h{160} \); the whole pole subtends \( \theta_1+\theta_2 \) with \( \tan\theta_2=\dfrac35 \).
2. \( \dfrac{\frac h{160}+\frac35}{1-\frac h{160}\cdot\frac35}=\dfrac h{40} \) gives \( h^2-200h+6400=0 \), so \( h=40 \) or \( 160 \). Of the options, \( 40 \) m. Answer: (b)

Teacher's Note:
1. The angle subtended by the whole pole is the sum of the two angles.