1. A vertical pole fixed to the ground is divided in the ratio 1 : 9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 15 m away from the base of the pole, what is the height of the pole ?
a) $$60\sqrt 5 \,{\text{m}}$$
b) $${\text{15}}\sqrt 5 \,{\text{m}}$$
c) $$15\sqrt 3 \,{\text{m}}$$
d) $${\text{60}}\sqrt 3 \,{\text{m}}$$
Explanation:
Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $$\theta $$
From "Angle Bisector Theorem", we have
$$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$
$$ \Rightarrow \frac{1}{9} = \frac{{15}}{{AC}}$$ [ BD : DC = 1 : 9 and AB = 15(given)]
$$ \Rightarrow AC = 15 \times 9\,{\text{m}}\,......\left( {eq:1} \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$
$$CB = \sqrt {A{C^2} - A{B^2}} $$ ( Pythagorean theorem)
$$ = \sqrt {{{\left( {15 \times 9} \right)}^2} - {{15}^2}} $$ [AC=15 × 9(eq : 1) and AB=15 m(given)]
$$\eqalign{ & = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} \cr & = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} \cr & = \sqrt {{{15}^2} \times 80} \cr & = \sqrt {{{15}^2} \times 16 \times 5} \cr & = 15 \times 4 \times \sqrt 5 \cr & = 60\sqrt 5 \,{\text{m}} \cr} $$
2. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
a) 26.28 km/hr
b) 32.42 km/hr
c) 24.22 km/hr
d) 31.25 km/hr
Explanation:
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
$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & 1 = \frac{{AB}}{{100}} \cr & AB = 100\,......\left( {eq:1} \right) \cr} $$
$$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$
$$ \frac{1}{{\sqrt 3 }} = \frac{{100}}{{BD}}$$ ( Substituted the value of AB from equation 1)
$$ BD = 100\sqrt 3 $$
$$\eqalign{ & CD = \left( {BD - BC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {100\sqrt 3 - 100} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 100\left( {\sqrt 3 - 1} \right) \cr} $$
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.
$$\eqalign{ & {\text{Required}}\,{\text{speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{10}} \cr & = 10\left( {1.73 - 1} \right) \cr & = 7.3\,{\text{meter/seconds}} \cr & = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = 26.28\,{\text{km/hr}} \cr} $$
3. An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other?
a) 381 m
b) 169 m
c) 254 m
d) 211 m
Explanation:
Let C and D be the position of the aeroplanes.
Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45°
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC, \cr & \tan {60^ \circ } = \frac{{CB}}{{AB}} \cr & \sqrt 3 = \frac{{900}}{{AB}} \cr & AB = \frac{{900}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{900 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{900\sqrt 3 }}{3} \cr & \,\,\,\,\,\,\,\,\,\, = 300\sqrt 3 \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABD, \cr & \tan {45^ \circ } = \frac{{DB}}{{AB}} \cr & 1 = \frac{{DB}}{{AB}} \cr & DB = AB = 300\sqrt 3 \cr & {\text{Required}}\,{\text{height}} = CD \cr & = \left( {CB - DB} \right) \cr & = \left( {900 - 300\sqrt 3 } \right) \cr & = \left( {900 - 300 \times 1.73} \right) \cr & = \left( {900 - 519} \right) \cr & = 381\,{\text{m}} \cr} $$
4. The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
a) 30°
b) 60°
c) 45°
d) None of these
Explanation:
Consider the diagram shown above where QR represents the tree and PQ represents its shadow
$$\eqalign{ & {\text{We}}\,{\text{have}},\,QR = PQ \cr & {\text{Let}}\,\angle QPR = \theta \cr} $$
$$\tan \theta = \frac{{QR}}{{PQ}} = 1$$ (since QR = PQ)
$$ \theta = {45^ \circ }$$
i.e., required angle of elevation = 45°
5. Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is the speed of the car?
a) $$24\sqrt 3 \,{\text{km/hr}}$$
b) $$\frac{{20\sqrt 3 }}{3}\,{\text{km/hr}}$$
c) $$\frac{{24}}{{\sqrt 3 }}\,{\text{km/hr}}$$
d) None of these
Explanation:
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60°
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABD, \cr & \tan {30^ \circ } = \frac{{BD}}{{BA}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{50}}{{BA}} \cr & BA = 50\sqrt 3 \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CBD, \cr & \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr & \sqrt 3 = \frac{{50}}{{BC}} \cr & \Rightarrow BC = \frac{{50}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 }}{3} \cr} $$
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 = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$
$$\eqalign{ & {\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}} \cr & = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = {\text{24}}\sqrt 3 \,{\text{km/hr}} \cr} $$.
6. The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree, is:
a) 30º
b) 45º
c) 60º
d) 90º
Explanation:
Let ∠ACB = θ
$$\frac{{AC}}{{AB}} = \cot \theta $$
$$\frac{{AC}}{{AB}} = \sqrt 3 $$
$$\cot {30^ \circ } = \sqrt 3 $$
θ = 30°
7. The angle of elevation of the top of a tower from a certain point is 30°. If the observed moves 20 m towards the tower, the angle of elevation the angle of elevation of top of the tower increases by 15°. The height of the tower is
a) 17.3 m
b) 21.9 m
c) 27.3 m
d) 30 m
Explanation:
Let AB be the tower and C and D be the points of observation.
Then, ∠ACB = 30°, ∠ADB = 45° and CD = 20m
Let AB = h then,
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AC = AB \times \sqrt 3 = h\sqrt 3 {\kern 1pt} {\text{And,}} \cr & \Rightarrow \frac{{AB}}{{AD}} = \tan 45^\circ = 1 \cr & \Rightarrow AD = AB = h \cr & \, \, \, \, \, CD = 20 \cr & \Rightarrow \left( {AC - AD} \right) = 20 \cr & \Rightarrow h\sqrt 3 - h = 20 \cr & h = \frac{{20}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = 10\left( {\sqrt 3 + 1} \right){\text{m}} \cr & = \left( {10 \times 2.73} \right){\text{m}} \cr & = 27.3 {\text{m}} \cr} $$
8. On the same side of tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is-
a) 63.5 m
b) 76.9 m
c) 86.7 m
d) 90 m
Explanation:
Let AB be the tower and C and D be the objects.
Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
$$\eqalign{ & \frac{{AB}}{{AD}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AD = \frac{{AB}}{{\sqrt 3 }} = \frac{{150}}{{\sqrt 3 }} \cr & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AC = AB = 150{\text{ m}} \cr & CD = \left( {AC - AD} \right) \cr & = \left( {150 - \frac{{150}}{{\sqrt 3 }}} \right){\text{m}} \cr & = \left[ {\frac{{150\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}} \right]{\text{m}} \cr & = 50\left( {3 - \sqrt 3 } \right){\text{m}} \cr & = \left( {50 \times 1.27} \right){\text{m}} \cr & = 63.5\,{\text{m}} \cr} $$
9. The angle of depression of a point situated at a distance of 70m from the base of a tower is 60°. The height of the tower is-
a) \[35\sqrt 3 {\text{ m}}\]
b) \[70\sqrt 3 {\text{ m}}\]
c) \[\frac{{70\sqrt 3 }}{3}{\text{ m}}\]
d) \[{\text{70 m}}\]
Explanation:
Length of the tower AB = h meter.
$$\eqalign{ & \angle DAC = \angle ACB = {60^ \circ } \cr & BC = 70{\text{ meter}} \cr & {\text{In }}\vartriangle {\text{ABC,}} \cr & {\text{tan }}{60^ \circ } = \frac{{AB}}{{BC}} \cr & \sqrt 3 = \frac{h}{{70}} \cr & h = 70\sqrt 3 {\text{ meter}} \cr} $$
10. A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower ?
a) 14 min. 35 sec.
b) 15 min. 49 sec.
c) 16 min. 23 sec.
d) 18 min. 5 sec.
Explanation:
Let AB be the tower and C and D be the two positions of the car.
Then, \[\angle ACB = {45^ \circ },\] \[\angle ADB = {30^ \circ }\]
Let, AB = h, CD = x and AC = y
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow \frac{h}{y} = 1 \cr & \Rightarrow y = h \cr & \frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }} \cr & x + y = \sqrt 3 h \cr & x = \left( {x + y - y} \right) \cr & \,\,\,\,\,\,\, = \sqrt 3 h - h \cr & \,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right) \cr} $$
Now, $$h\left( {\sqrt 3 - 1} \right)$$ is covered in 12 min.
So, h will be covered in→
$$\eqalign{ & \left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right] \cr & = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min \cr & = \left( {\frac{{1200}}{{73}}} \right)\min \cr & = 16\min ,\,\,23\sec \cr} $$