Properties of Triangle Questions — JEE Mathematics

Q1

A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the centroid of this triangle is :

A

\left( {{1 \over 3},2} \right)

B

\left( {{1 \over 3},{5 \over 3}} \right)

C

\left( {1,{7 \over 3}} \right)

D

\left( {{1 \over 3},1} \right)

Correct Answer

Option A

Solution

Centroid =

\left( {{{1 + 3 - 3} \over 3},{{2 + 0 + 4} \over 3}} \right) = \left( {{1 \over 3},2} \right)

Q2

Two poles, AB of length a metres and CD of length a + b (b

\ne

a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan

\angle

ACB =

{1 \over 2}

, then :

A x2 + 2(a + 2b)x

-

b(a + b) = 0

B x2 + 2(a + 2b)x + a(a + b) = 0

C x2

-

2ax + b(a + b) = 0

D x2

-

2ax + a(a + b) = 0

Correct Answer

Option C

Solution

\tan \theta = {1 \over 2}

\tan (\theta + \alpha ) = {x \over b},\tan \alpha = {x \over {a + b}}

\Rightarrow {{{1 \over 2} + {x \over {a + b}}} \over {1 - {1 \over 2} \times {x \over {a + b}}}} = {x \over b}

\Rightarrow {x^2} - 2ax + ab + {b^2} = 0

Q3

If the angles of elevation of the top of a tower from three collinear points

A, B

and

C,

on a line leading to the foot of the tower, are

{30^ \circ }

,

{45^ \circ }

and

{60^ \circ }

respectively, then the ratio,

AB:BC,

is :

A

1:\sqrt 3

B

2:3

C

\sqrt 3 :1

D

\sqrt 3 :\sqrt 2

Correct Answer

Option C

Solution

As

PB

bisects

\angle APC,

therefore

AB

:

BC

=PA:PC

Also in

\Delta APQ,\sin {30^ \circ } = {h \over {PA}} \Rightarrow PA = 2h

and in

\Delta CPQ,

\sin {60^ \circ } = {h \over {PC}} \Rightarrow PC = {{2h} \over {\sqrt 3 }}

$\therefore$

AB:BC = 2h:{{2h} \over {\sqrt 3 }} = \sqrt 3 :1

Q4

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is

{60^ \circ }

and when he retires

40

meters away from the tree the angle of elevation becomes

{30^ \circ }

. The breadth of the river is :

A

60\,\,m

B

30\,\,m

C

40\,\,m

D

20\,\,m

Correct Answer

Option D

Solution

From the figure

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

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

\tan {30^ \circ } = {y \over {x + 40}}

\Rightarrow y = {{x + 40} \over {\sqrt 3 }}........\left( 2 \right)

From

(1)

and

(2),

\sqrt 3 x = {{x + 40} \over {\sqrt 3 }} \Rightarrow x = 20m

Q5

A tower stands at the centre of a circular park.

A

and

B

are two points on the boundary of the park such that

AB(=a)

subtends an angle of

{60^ \circ }

at the foot of the tower, and the angle of elevation of the top of the tower from

A

or

B

is

{30^ \circ }

. The height of the tower is :

A

a/\sqrt 3

B

a\sqrt 3

C

2a/\sqrt 3

D

2a\sqrt 3

Correct Answer

Option A

Solution

In the

\Delta AOB,\,\,\angle AOB = {60^ \circ },

and

\angle OBA = \angle OAB

(since

OA=OB=AB

radius of same circle). $\therefore$

\Delta AOB

is a equilateral triangle. Let the height of tower is

h

Given distance between two points

A

&

B

lie on boundary of circular park, subtends an angle of

{60^ \circ }

at the foot of the tower is

AB

i.e.

AB

=a.

A tower

OC

stands at the center of a circular park. Angle of elevation of the top of the tower from

A

and

B

is

{30^ \circ }

. In

\Delta OAC\,\,\tan {30^ \circ } = {h \over a}

\Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}

Q6

AB

is a vertical pole with

B

at the ground level and

A

at the top.

A

man finds that the angle of elevation of the point

A

from a certain point

C

on the ground is

{60^ \circ }

. He moves away from the pole along the line

BC

to a point

D

such that

CD=7

m. From

D

the angle of elevation of the point

A

is

{45^ \circ }

. Then the height of the pole is :

A

{{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 - 1} }}m

B

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

C

{{7\sqrt 3 } \over 2}\left( {\sqrt {3 } - 1 } \right)m

D

{{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 + 1} }}m

Correct Answer

Option B

Solution

In

\Delta ABC

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

\Rightarrow x = {h \over {\sqrt 3 }}

In

\Delta ABD{h \over {x + 7}}

= \tan {45^ \circ } = 1

\Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7

\Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}

\Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3 + 1\,m} \right)

Q7

ABCD

is a trapezium such that

AB

and

CD

are parallel and

BC \bot CD.

If

\angle ADB = \theta ,\,BC = p

and

CD = q,

then AB is equal to:

A

{{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {p\cos \theta + q\sin \theta }}

B

{{{p^2} + {q^2}\cos \theta } \over {p\cos \theta + q\sin \theta }}

C

{{{p^2} + {q^2}} \over {{p^2}\cos \theta + {q^2}\sin \theta }}

D

{{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {{{\left( {p\cos \theta + q\sin \theta } \right)}^2}}}

Correct Answer

Option A

Solution

From Sine Rule

{{AB} \over {\sin \theta }} = {{\sqrt {{p^2} + {q^2}} } \over {\sin \left( {\pi - \left( {\theta + \alpha } \right)} \right)}}

AB = {{\sqrt {{p^2} + {q^2}} \sin \theta } \over {\sin \theta \cos \alpha + \cos \theta \sin \alpha }}

= {{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {q\sin \theta + p\cos \theta }}

(As

\cos \alpha = {q \over {\sqrt {{p^2} + {q^2}} }}

and

\sin \alpha = {p \over {\sqrt {{p^2} + {q^2}} }}

)

Q8

A bird is sitting on the top of a vertical pole

20

m high and its elevation from a point

O

on the ground is

{45^ \circ }

. It files off horizontally straight away from the point

O

. After one second, the elevation of the bird from

O

is reduced to

{30^ \circ }

. Then the speed (in m/s) of the bird is :

A

20\sqrt 2

B

20\left( {\sqrt 3 - 1} \right)

C

40\left( {\sqrt 2 - 1} \right)

D

40\left( {\sqrt 3 - \sqrt 2 } \right)

Correct Answer

Option B

Solution

Let the speed be

y

m/sec

. Let

AC

be the vertical pole of height

20

m.

Let

O

be the point on the ground such that

\angle AOC = {45^ \circ }

Let

OC = x

Time

t=1

s

From

\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right)

and from

\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right)

From

(i)

and

(ii),

we have

x=20

and

{1 \over {\sqrt 3 }} = {{20} \over {x + y}}

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

\Rightarrow 20 + y = 20\sqrt 3

So,

y = 20\left( {\sqrt 3 - 1} \right)\,\,i.e.,

speed

= 20\left( {\sqrt 3 - 1} \right)m/s

Q9

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60o. Then the time taken (in minutes) by him, from B to reach the pillar, is :

A 6

B 10

C 20

D 5

Correct Answer

Option D

Solution

According to given information, we have the following figure Now, from $\triangle A C D$ and $\triangle B C D$ , we have $\tan 30^{\circ}=\dfrac{h}{x+y}$ and $\tan 60^{\circ}=\dfrac{h}{y}$ $\Rightarrow h=\dfrac{x+y}{\sqrt{3}}\quad...(i)$ and $h=\sqrt{3} y\quad...(ii)$ From Eqs. (i) and (ii), $\dfrac{x+y}{\sqrt{3}}=\sqrt{3} y \Rightarrow x+y=3 y$ $\Rightarrow x-2 y=0 \Rightarrow y=\dfrac{x}{2}$ $\because$ Speed is uniform.

$\therefore$ Distance $y$ will be cover in $5 \mathrm{~min}$ .

$\because$ Distance $x$ covered in $10 \mathrm{~min}$ .

$\therefore$ Distance $\dfrac{x}{2}$ will be cover in $5 \mathrm{~min}$ .

Q10

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If

\angle

BPC =

\beta

, then tan

\beta

is equal to:

A

{1 \over 4}

B

{2 \over 9}

C

{4 \over 9}

D

{6 \over 7}

Correct Answer

Option B

Solution

Let the height of tower

AB = x

and

LCPA = \propto

From the diagram you can see,

\tan \left( { \propto + \beta } \right) = {x \over {2x}} = {1 \over 2}

we know,

\tan \left( { \propto + \beta } \right) = {{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }}

\therefore\,\,\,

{{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }} = {1 \over 2}....\left( 1 \right)

From the diagram,

\tan \widehat \propto = {{x/2} \over {2x}} = {1 \over 4}......\left( 2 \right)

Putting value of

\tan \propto

in eq

(1)

,

{{{1 \over 4} + \tan \beta } \over {1 - {1 \over 4}\tan \beta }} = {1 \over 2}

\Rightarrow 1 - {1 \over 4}\tan \beta

= {1 \over 2} + 2\tan \beta

\Rightarrow {{9\tan \beta } \over 4} = {1 \over 2}

\Rightarrow \tan \beta = {2 \over 9}