Properties of Triangle — Page 2 | JEE Mathematics

Q11

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45

^\circ

, 30

^\circ

and 30

^\circ

, then the height of the tower (in m) is :

A

50\sqrt 2

B 100

C 50

D

100\sqrt 3

Correct Answer

Option B

Solution

Let height of tower

TM = h

From triangle

PMT,

\tan {45^o} = {h \over {PM}}

\Rightarrow \,\,\,PM = h

From triangle

TRM,

\tan {30^o} = {{TM} \over {RM}}

\Rightarrow \,\,\,RM = \sqrt3 h

From triangle.

PMR,

P{M^2} + M{R^2} = {\left( {200} \right)^2}

\Rightarrow \,\,\,h{}^2 + {\left( {\sqrt 3 h} \right)^2} = {200^2}

\Rightarrow \,\,\,4{h^2} = {200^2}

\Rightarrow \,\,\,{h^2} = {{{{200}^2}} \over 4}

\Rightarrow \,\,\,h = {{200} \over 2} = 100\,m

\therefore\,\,\,

height of tower

=100

m.

Q12

An aeroplane flying at a constant speed, parallel to the horizontal ground,

\sqrt 3

kmabove it, is obsered at an elevation of

{60^o}

from a point on the ground. If, after five seconds, its elevation from the same point, is

{30^o}

, then the speed (in km / hr) of the aeroplane, is :

A 1500

B 1440

C 750

D 720

Correct Answer

Option B

Solution

For

\Delta

OA, A, OA1 =

{{\sqrt 3 } \over {\tan {{60}^o}}}

= 1 km For

\Delta

OB1, B, OB1 =

{{\sqrt 3 } \over {\tan {{30}^o}}}

= 3km As, a distance of 3 $-$ 1 = 2 km is convered in 5 seconds. Therefore the speed of the plane is

{{2 \times 3600} \over 5}

= 1440 km/hr

Q13

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45o from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30o, then the distance (in m) of the foot of the tower from the point A is :

A

15\left( {1 + \sqrt 3 } \right)

B

15\left( {3 - \sqrt 3 } \right)

C

15\left( {3 + \sqrt 3 } \right)

D

15\left( {5 - \sqrt 3 } \right)

Correct Answer

Option C

Solution

{x \over d} = \tan {30^o} \Rightarrow {x \over d} = {1 \over {\sqrt 3 }} \Rightarrow d = \sqrt 3 x

and

{{x + 30} \over d} = \tan {45^o}

$\Rightarrow$ d = x + 30

d = {d \over {\sqrt 3 }} + 30 \Rightarrow \left( {1 - {1 \over {\sqrt 3 }}} \right)d = 30

\Rightarrow d = {{30\sqrt 3 } \over {\sqrt 3 - 1}} \Rightarrow d = {{30\sqrt 3 (\sqrt 3 + 1)} \over 2} = 15\sqrt 3 (\sqrt 3 + 1)

$\Rightarrow$ d = 15 ( 3 +

\sqrt 3

)

Q14

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot–1 (3

\sqrt 2

) and cosec–1 (2

\sqrt 2

) respectively, then the height of the tower (in metres) is :

A

{{100} \over {3\sqrt 3 }}

B 25

C 20

D 10

\sqrt 5

Correct Answer

Option C

Solution

\Delta

APM

{h \over {AM}} = {1 \over {3\sqrt 2 }}

\Delta

BPM

{h \over {BM}} = {1 \over {\sqrt 7 }}

\Delta

ABM AM2 + MB2 = (100)2 ' $\Rightarrow$ 18h2 + 7h2 = 100 × 100 $\Rightarrow$ h2 = 4 × 100 $\Rightarrow$ h = 20

Q15

Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-

A

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

B

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

C

10\left( {\sqrt3 - 1 } \right)

D

5\left( {\sqrt3 + 1 } \right)

Correct Answer

Option A

Solution

\tan {15^o} = {5 \over d} \Rightarrow d = {5 \over {\tan {{15}^o}}}

\Rightarrow {{5\left( {\sqrt 3 + 1} \right)} \over {\sqrt 3 - 1}} \Rightarrow {{5\left( {4 + 2\sqrt 3 } \right)} \over 2}

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

Q16

Two vertical poles of heights, 20 m and 80 m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

A 12

B 16

C 15

D 18

Correct Answer

Option B

Solution

From triangle BCD, tan $\alpha$ =

{{80} \over x}

and from triangle BFE, tan $\alpha$ =

{{h} \over y}

$\therefore$

{{80} \over x}

=

{{h} \over y}

$\Rightarrow$

y = {{hx} \over {80}}

........ (1) From triangle ABC, tan $\beta$ =

{{20} \over x}

and from triangle EFC, tan $\beta$ =

{{h} \over {x-y}}

$\therefore$

{{20} \over x}

=

{{h} \over {x-y}}

$\Rightarrow$

x-y = {{hx} \over {20}}

........ (2) By adding equation (1) and (2) we get, x =

{{hx} \over {80}}

+

{{hx} \over {20}}

$\Rightarrow$ 1 =

{{h} \over {80}}

+

{{h} \over {20}}

$\Rightarrow$ h = 16 m $\therefore$ Height of intersection point is 16 m

Q17

The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60°. Then the height (in km) of the summit from the ground is :

A

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

B

{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}

C

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

D

{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}

Correct Answer

Option A

Solution

In

\Delta

CDF sin 30o =

{z \over 1}

$\Rightarrow$ z =

{1 \over 2}

km cos 30o =

{Y \over 1}

$\Rightarrow$ Y =

{{\sqrt 3 } \over 2}

km Now in

\Delta

ABC, tan 45o =

{h \over {X + Y}}

$\Rightarrow$ h = X + Y $\Rightarrow$ X = h -

{{\sqrt 3 } \over 2}

Now in

\Delta

BDE tan 60o =

{{h - z} \over X}

$\Rightarrow$

{\sqrt 3 }

X = h -

{1 \over 2}

$\Rightarrow$

\sqrt 3 \left( {h - {{\sqrt 3 } \over 2}} \right)

= h -

{1 \over 2}

$\Rightarrow$ h =

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

km

Q18

The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30°. If the angle of depression of the image of C in the lake from the point P is 60°,then PC (in m) is equal to :

A

200\sqrt 3

B 400

C 100

D

400\sqrt 3

Correct Answer

Option B

Solution

Let PA = x For

\Delta

APC

AC = {{PA} \over {\sqrt 3 }} = {x \over {\sqrt 3 }}

A{C^1} = AB + B{C^1}

A{C^1} = 200 + {x \over {\sqrt 3 }}

From

\Delta {C^1}PA,

A{C^1} = \sqrt 3 PA

\Rightarrow \left( {200 + {x \over {\sqrt 3 }}} \right) = \sqrt 3 x

\Rightarrow x = (200)(\sqrt 3 )

From

\Delta APC,

sin 30o =

{{{x \over {\sqrt 3 }}} \over {PC}}

$\Rightarrow$

PC = {{2x} \over {\sqrt 3 }}

\Rightarrow PC = 400

Q19

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is :

A 30

B 25

C 20

\sqrt 3

D 25

\sqrt 3

Correct Answer

Option D

Solution

$\therefore$

\tan (90 - \theta ) = {x \over {75}}

and

\tan \theta = {{3x} \over {75}}

As

\cot \theta .\tan \theta = 1

$\therefore$

{x \over {75}}.{{3x} \over {75}} = 1

\Rightarrow x = 25\sqrt 3

Q20

The angle of elevation of a jet plane from a point A on the ground is 60

^\circ

. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30

^\circ

. If the jet plane is flying at a constant height, then its height is :

A

3600\sqrt 3

m

B

1200\sqrt 3

m

C

1800\sqrt 3

m

D

2400\sqrt 3

m

Correct Answer

Option B

Solution

v = 432 \times {{1000} \over {60 \times 60}}

m/sec = 120 m/sec Distance AB = v $\times$ 20 = 2400 meter In

\Delta

PAC

\tan 60^\circ = {h \over {PC}} \Rightarrow PC = {h \over {\sqrt 3 }}

In

\Delta

PBD

\tan 30^\circ = {h \over {PD}} \Rightarrow PD = \sqrt 3 h

PD = PC + CD

\sqrt 3 h = {h \over {\sqrt 3 }} + 2400 \Rightarrow {{2h} \over {\sqrt 3 }} = 2400

h = 1200\sqrt 3

meter