Properties of Triangle — Page 3 | JEE Mathematics

Q21

A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30

^\circ

(Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45

^\circ

. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

A

10(\sqrt 3 + 1)

B

10(\sqrt 3 - 1)

C 10

D

10\sqrt 3

Correct Answer

Option A

Solution

{h \over {x + y}} = \tan 30^\circ

x + y = \sqrt 3 h

...... (1) Also,

{h \over y} = \tan 45^\circ

h = y

..... (2) put in (1)

x + y = \sqrt 3 y

x = \left( {\sqrt 3 - 1} \right)y

{x \over {20}} = 'v'

speed $\therefore$ time taken to reach Foot from B

= {y \over V}

= {x \over {\left( {\sqrt 3 - 1} \right).x}} \times 20

= 10\left( {\sqrt 3 + 1} \right)

Q22

A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be

{\pi \over 3}

. If the radius of the circumcircle of

\Delta

ABC is 2, then the height of the pole is equal to :

A

{{1 \over {\sqrt 3 }}}

B 2

{\sqrt 3 }

C

{\sqrt 3 }

D

{{{2\sqrt 3 } \over 3}}

Correct Answer

Option B

Solution

Let PD = h, R = 2 As angle of elevation of top of pole from A, B, C are equal. So D must be circumcentre of

\Delta

ABC

\tan \left( {{\pi \over 3}} \right) = {{PD} \over R} = {h \over R}

h = R\tan \left( {{\pi \over 3}} \right) = 2\sqrt 3

Q23

Let in a right angled triangle, the smallest angle be

\theta

. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin

\theta

is equal to :

A

{{\sqrt 5 + 1} \over 4}

B

{{\sqrt 5 - 1} \over 2}

C

{{\sqrt 2 - 1} \over 2}

D

{{\sqrt 5 - 1} \over 4}

Correct Answer

Option B

Solution

Let a

\Delta

ABC having C = 90

^\circ

and A = $\theta$

{{\sin \theta } \over a} = {{\cos \theta } \over b} = {1 \over c}

..... (i) Also for triangle of reciprocals

\cos A = {{{{\left( {{1 \over c}} \right)}^2} + {{\left( {{1 \over b}} \right)}^2} - {{\left( {{1 \over a}} \right)}^2}} \over {2\left( {{1 \over c}} \right)\left( {{1 \over b}} \right)}}

{1 \over {{c^2}}} + {1 \over {{{(c\cos \theta )}^2}}} = {1 \over {{{(c\sin \theta )}^2}}}

\Rightarrow 1 + {\sec ^2}\theta = \cos e{c^2}\theta

\Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }}

\Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {{{\sin }^2}2\theta }}

\Rightarrow 1 - {\cos ^2}2\theta = 4\cos 2\theta

{\cos ^2}2\theta + 4\cos 2\theta - 1 = 0

\cos 2\theta = {{ - 4 \pm \sqrt {16 + 4} } \over 2}

\cos 2\theta = - 2 \pm \sqrt 5

\cos 2\theta = \sqrt 5 - 2 = 1 - 2{\sin ^2}\theta

\Rightarrow 2{\sin ^2}\theta = 3 - \sqrt 5

\Rightarrow {\sin ^2}\theta = {{3 - \sqrt 5 } \over 2}

\Rightarrow \sin \theta = {{\sqrt 5 - 1} \over 2}

Q24

A spherical gas balloon of radius 16 meter subtends an angle 60

^\circ

at the eye of the observer A while the angle of elevation of its center from the eye of A is 75

^\circ

. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :

A

8(2 + 2\sqrt 3 + \sqrt 2 )

B

8(\sqrt 6 + \sqrt 2 + 2)

C

8(\sqrt 2 + 2 + \sqrt 3 )

D

8(\sqrt 6 - \sqrt 2 + 2)

Correct Answer

Option B

Solution

O $\to$ centre of sphere P, Q $\to$ point of contact of tangents from A Let T be top most point of balloon & R be foot of perpendicular from O to ground.

From triangle OAP, OA = 16cosec30

^\circ

= 32 From triangle ABO, OR = OA sin75

^\circ

=

32{{\left( {\sqrt 3 + 1} \right)} \over {2\sqrt 2 }}

So level of top most point = OR + OT

= 8\left( {\sqrt 6 + \sqrt 2 + 2} \right)

Q25

A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :

A 12

\sqrt {15}

B 12

\sqrt {10}

C 8

\sqrt {10}

D 6

\sqrt {10}

Correct Answer

Option B

Solution

Let height of pole = 10

l

\tan \alpha = {{3l} \over {18}} = {l \over 6}

\tan 2\alpha = {{10l} \over {18}}

{{2\tan \alpha } \over {1 - {{\tan }^2}\alpha }} = {{10l} \over {18}}

use

\tan \alpha = {l \over 6} \Rightarrow l = \sqrt {{{72} \over 5}}

height of pole =

10l = 12\sqrt {10}

Q26

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60

^\circ

. The pole subtends an angle 30

^\circ

at the top of the tower. Then the height of the tower is :

A

15\sqrt 3

B

20\sqrt 3

C 20 +

10\sqrt 3

D 30

Correct Answer

Option D

Solution

Here AB is a tower and CD is a pole. In triangle ABC,

\tan 60^\circ = {{AB} \over {AC}} = {{20 + h} \over x}

...... (1) In triangle BED,

\tan 30^\circ = {h \over x}

...... (2) Divide equation (1) by equation (2), we get

{{\tan 60^\circ } \over {\tan 30^\circ }} = {{20 + h} \over x} \times {x \over h}

\Rightarrow {{\sqrt 3 } \over {{1 \over {\sqrt 3 }}}} = {{20 + h} \over h}

\Rightarrow 3 = {{20 + h} \over h}

\Rightarrow 3h = 20 + h

\Rightarrow h = 10\,m

$\therefore$ Height of tower

= 20 + 10 = 30\,m

Q27

Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let

{\pi \over 8}

and

\theta

be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan2

\theta

is equal to

A

{{3 - 2\sqrt 2 } \over 2}

B

{{3 + \sqrt 2 } \over 2}

C

{{3 - 2\sqrt 2 } \over 4}

D

{{3 - \sqrt 2 } \over 4}

Correct Answer

Option C

Solution

{l \over {80}} = \tan \theta

..... (i)

{{2l} \over {80}} = \tan {\pi \over 8}

..... (ii) From (i) and (ii)

{1 \over 2} = {{\tan \theta } \over {\tan {\pi \over 8}}} \Rightarrow {\tan ^2}\theta = {1 \over 4}{\tan ^2}{\pi \over 8}

\Rightarrow {\tan ^2}\theta = {{\sqrt 2 - 1} \over {4(\sqrt 2 + 1)}} = {{3 - 2\sqrt 2 } \over 4}

Q28

A tower PQ stands on a horizontal ground with base

Q

on the ground. The point

R

divides the tower in two parts such that

Q R=15 \mathrm{~m}

. If from a point

A

on the ground the angle of elevation of

R

is

60^{\circ}

and the part

P R

of the tower subtends an angle of

15^{\circ}

at

A

, then the height of the tower is :

A

5(2 \sqrt{3}+3) \,\mathrm{m}

B

5(\sqrt{3}+3) \,\mathrm{m}

C

10(\sqrt{3}+1) \,\mathrm{m}

D

10(2 \sqrt{3}+1) \,\mathrm{m}

Correct Answer

Option A

Solution

For

\Delta

AQR,

\tan 60^\circ = {{15} \over x}

....... (1) From

\Delta

AQP,

\tan 75^\circ = {h \over x}

\Rightarrow \left( {2 + \sqrt 3 } \right) = {h \over x}

[ $\because$

\tan 75^\circ = 2 + \sqrt 3

]

\Rightarrow h = \left( {2 + \sqrt 3 } \right)x

= \left( {2 + \sqrt 3 } \right){{15} \over {\sqrt 3 }}

[From (1)]

= \left( {2 + \sqrt 3 } \right) \times {{15\sqrt 3 } \over 3}

= \left( {2 + \sqrt 3 } \right) \times 5\sqrt 3

= 5\left( {2\sqrt 3 + 3} \right)

m

Q29

Let a vertical tower

A B

of height

2 h

stands on a horizontal ground. Let from a point

P%

on the ground a man can see upto height

h

of the tower with an angle of elevation

2 \alpha

. When from

P

, he moves a distance

d

in the direction of

\overrightarrow{A P}

, he can see the top

B

of the tower with an angle of elevation

\alpha

. If

d=\sqrt{7} h

, then

\tan \alpha

is equal to

A

\sqrt{5}-2

B

\sqrt{3}-1

C

\sqrt{7}-2

D

\sqrt{7}-\sqrt{3}

Correct Answer

Option C

Solution

\Delta{APM}

gives

\tan 2\alpha = {h \over x}

..... (i)

\Delta{AQB}

gives

\tan 2\alpha = {{2h} \over {x + d}} = {{2h} \over {x + h\sqrt 7 }}

...... (ii) From (i) and (ii)

\tan 2\alpha = {{2\,.\,\tan 2\alpha } \over {1 + \sqrt 7 \,.\,\tan 2\alpha }}

Let

t = \tan \alpha

\Rightarrow t = {{2{{2t} \over {1 - {t^2}}}} \over {1 + \sqrt 7 \,.\,{{2t} \over {1 - {t^2}}}}}

\Rightarrow {t^2} - 2\sqrt 7 t + 3 = 0

t = \sqrt 7 - 2

Q30

The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is

45^{\circ}

. Let R be a point on AQ and from a point B, vertically above

\mathrm{R}

, the angle of elevation of

\mathrm{P}

is

60^{\circ}

. If

\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d}

and the area of the trapezium

\mathrm{PQRB}

is

\alpha

, then the ordered pair

(\mathrm{d}, \alpha)

is :

A

(10(\sqrt{3}-1), 25)

B

\left(10(\sqrt{3}-1), \dfrac{25}{2}\right)

C

(10(\sqrt{3}+1), 25)

D

\left(10(\sqrt{3}+1), \dfrac{25}{2}\right)

Correct Answer

Option A

Solution

Let

BR = x

{x \over d} = {1 \over 2} \Rightarrow x = {d \over 2}

{{10 - x} \over {10 - x\sqrt 3 }} = \sqrt 3 \Rightarrow 10 - x = 10\sqrt 3 - 3x

2x = 10(\sqrt 3 - 1)

x = 5(\sqrt 3 - 1)

d = 2x = 10(\sqrt 3 - 1)

\alpha = {1 \over 2}(x + 10)(10 - x\sqrt 3 )=

Area (PQRB)

= {1 \over 2}\left( {5\sqrt 3 - 5 + 10} \right)\left( {10 - 5\sqrt 3 (\sqrt 3 - 1)} \right)

= {1 \over 2}\left( {5\sqrt 3 + 5} \right)\left( {10 - 15 + 5\sqrt 3 } \right) - {1 \over 2}(75 - 25) = 25