Properties of Triangle — Page 4 | JEE Mathematics

Q31

A horizontal park is in the shape of a triangle

\mathrm{OAB}

with

\mathrm{AB}=16

. A vertical lamp post

\mathrm{OP}

is erected at the point

\mathrm{O}

such that

\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ}

and

\angle \mathrm{PCO}=45^{\circ}

, where

\mathrm{C}

is the midpoint of

\mathrm{AB}

. Then

(\mathrm{OP})^{2}

is equal to :

A

\dfrac{32}{\sqrt{3}}(\sqrt{3}-1)

B

\dfrac{32}{\sqrt{3}}(2-\sqrt{3})

C

\dfrac{16}{\sqrt{3}}(\sqrt{3}-1)

D

\dfrac{16}{\sqrt{3}}(2-\sqrt{3})

Correct Answer

Option B

Solution

OP = OA\tan 15 = OB\tan 15

...... (i)

OP = OC\tan 45 \Rightarrow OP = OC

...... (ii)

OA = OB

....... (iii)

O{C^2} + {8^2} = O{A^2}

O{P^2} + 64 = O{P^2}{\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right)^2}

64 = O{P^2}\left[ {{{{{\left( {\sqrt 3 + 1} \right)}^2} - {{\left( {\sqrt {3 - 1} } \right)}^2}} \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right]

= O{P^2}\left( {{{4\sqrt 3 } \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right)

O{P^2} = {{64{{\left( {\sqrt 3 - 1} \right)}^2}} \over {4\sqrt 3 }} = {{32} \over {\sqrt 3 }}\left( {2 - \sqrt 3 } \right)

Q32

The angle of elevation of the top of a tower from a point A due north of it is

\alpha

and from a point B at a distance of 9 units due west of A is

\cos ^{-1}\left(\dfrac{3}{\sqrt{13}}\right)

. If the distance of the point B from the tower is 15 units, then

\cot \alpha

is equal to :

A

\dfrac{6}{5}

B

\dfrac{9}{5}

C

\dfrac{4}{3}

D

\dfrac{7}{3}

Correct Answer

Option A

Solution

NA = \sqrt {{{15}^2} - {9^2}} = 12

{h \over {15}} = \tan \theta = {2 \over 3}

h = 10

units

\cot \alpha = {{12} \over {10}} = {6 \over 5}

Q33

The angle of elevation of the top

\mathrm{P}

of a tower from the feet of one person standing due South of the tower is

45^{\circ}

and from the feet of another person standing due west of the tower is

30^{\circ}

. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to

A 10

B

\dfrac{5}{2} \sqrt{5}

C

5 \sqrt{5}

D 5

Correct Answer

Option A

Solution

Let's denote the person standing due south as S and the one standing due west as W.

Also, let the tower be at point T.

From person S's perspective, we have a right triangle $\triangle SPT$ .

The height of the tower PT is given as 5m, which is the opposite side for angle S.

The angle at S is $45^\circ$ .

From the tangent trigonometric ratio, we have : $\tan{45^\circ} = \dfrac{PT}{ST} = \dfrac{5}{ST}$ Since $\tan{45^\circ}=1$ , we get $ST = 5$ m.

Similarly, from person W's perspective, we have a right triangle $\triangle WPT$ .

The angle at W is $30^\circ$ .

From the tangent trigonometric ratio, we have: $\tan{30^\circ} = \dfrac{PT}{WT} = \dfrac{5}{WT}$ Since $\tan{30^\circ}=\dfrac{1}{\sqrt{3}}$ , we get $WT = 5\sqrt{3}$ m.

Since S and W are perpendicular to each other (one is due south and the other is due west), $\triangle SWT$ forms a right triangle.

We can find $SW$ (the distance between the two people) using the Pythagorean theorem: $SW^2 = ST^2 + WT^2 = 5^2 + (5\sqrt{3})^2 = 25 + 75 = 100$ So, $SW = \sqrt{100} = 10$ m.

Hence, the correct answer is 10 meters, which corresponds to Option A.

Q34

For a regular polygon, let

r

and

R

be the radii of the inscribed and the circumscribed circles. A

false

statement among the following is :

A There is a regular polygon with

{r \over R} = {1 \over {\sqrt 2 }}

B There is a regular polygon with

{r \over R} = {2 \over 3}

C There is a regular polygon with

{r \over R} = {{\sqrt 3 } \over 2}

D There is a regular polygon with

{r \over R} = {1 \over 2}

Correct Answer

Option B

Solution

If

O

is center of polygon and

AB

is one of the side, then by figure

\cos {\pi \over n} = {r \over R}

\Rightarrow {r \over R} = {1 \over 2},{1 \over {\sqrt 2 }},{{\sqrt 3 } \over 2}\,\,for

n = 3,4,6

respectively.

Q35

From the top

\mathrm{A}

of a vertical wall

\mathrm{AB}

of height

30 \mathrm{~m}

, the angles of depression of the top

\mathrm{P}

and bottom

\mathrm{Q}

of a vertical tower

\mathrm{PQ}

are

15^{\circ}

and

60^{\circ}

respectively,

\mathrm{B}

and

\mathrm{Q}

are on the same horizontal level. If

\mathrm{C}

is a point on

\mathrm{AB}

such that

\mathrm{CB}=\mathrm{PQ}

, then the area (in

\mathrm{m}^{2}

) of the quadrilateral

\mathrm{BCPQ}

is equal to :

A

200(3-\sqrt{3})

B

300(\sqrt{3}-1)

C

300(\sqrt{3}+1)

D

600(\sqrt{3}-1)

Correct Answer

Option D

Solution

Given, $A B$ be a vertical wall of height $30 \mathrm{~m}$ and $P Q$ be a vertical tower.

Such that $\angle B Q A=\angle T A Q=60^{\circ}$ (Alternate angles) and $\angle C P A=\angle T A P=15^{\circ}$ Now, $\tan 60^{\circ}=\dfrac{A B}{B Q}=\dfrac{30}{B Q}$

\begin{array}{ll} &\Rightarrow \sqrt{3}=\frac{30}{B Q} \\\\ &\Rightarrow B Q=\frac{30}{\sqrt{3}}=10 \sqrt{3} \end{array}

\begin{aligned} & \text{ and } \tan 15^{\circ}=\frac{A C}{P C} \\\\ & \Rightarrow 2-\sqrt{3}=\frac{A C}{10 \sqrt{3}} (\because P C=B Q) \\\\ & \Rightarrow A C=10 \sqrt{3}(2-\sqrt{3})=20 \sqrt{3}-30 \\\\ & \therefore B C=A B-A C=30-(20 \sqrt{3}-30)=60-20 \sqrt{3} \end{aligned}

$\therefore$ Area of quadrilateral $B C P Q$

\begin{aligned} & =B Q \times B C=10 \sqrt{3} \times(60-20 \sqrt{3}) \\\\ & =10(60 \sqrt{3}-60) \\\\ & =600(\sqrt{3}-1) \mathrm{m}^2 \end{aligned}

Q36

The sides of a triangle are

3x + 4y,

4x + 3y

and

5x + 5y

where

x

,

y>0

then the triangle is :

A right angled

B obtuse angled

C equilateral

D none of these

Correct Answer

Option B

Solution

Let

\,\,\,\,a = 3x + 4y,b = 4x + 3y

and

c = 5x + 5y

as

\,\,\,\,x,y > 0,c = 5x + 5y

is the largest side $\therefore$

C

is the largest angle. Now

\cos \,C = {{{{\left( {3x + 4y} \right)}^2} + {{\left( {4x + 3y} \right)}^3} - {{\left( {5x + 5y} \right)}^2}} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}}

= {{ - 2xy} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} < 0

$\therefore$

C

is obtuse angle

\Rightarrow \Delta ABC

is obtuse angled

Q37

In a triangle with sides

a, b, c,

{r_1} > {r_2} > {r_3}

(which are the ex-radii) then :

A

a>b>c

B

a < b < c

C

a > b

and

b < c

D

a < b

and

b > c

Correct Answer

Option A

Solution

{r_1} > {r_2} > {r_3}

\Rightarrow {\Delta \over {s - a}} > {\Delta \over {s - b}} > {\Delta \over {s - c}};

\Rightarrow s - a < s - b < s - c

\Rightarrow - a < - b < - c

\Rightarrow a > b > c

Q38

The sum of the radii of inscribed and circumscribed circles for an

n

sided regular polygon of side

a,

is :

A

{a \over 4}\cot \left( {{\pi \over {2n}}} \right)

B

a\cot \left( {{\pi \over {n}}} \right)

C

{a \over 2}\cot \left( {{\pi \over {2n}}} \right)

D

a\cot \left( {{\pi \over {2n}}} \right)

Correct Answer

Option C

Solution

\tan \left( {{\pi \over n}} \right) = {a \over {2r}};\,\,\sin \left( {{\pi \over n}} \right) = {a \over {2R}}

r + R = {a \over 2}\left[ {\cot {\pi \over n} + \cos ec{\pi \over n}} \right]

= {a \over 2}\left[ {{{\cos {\pi \over n} + 1} \over {\sin {\pi \over n}}}} \right]

= {a \over 2}\left[ {{{2{{\cos }^2}{\pi \over {2n}}} \over {2\sin {\pi \over {2n}}\cos {\pi \over {2n}}}}} \right]

= {a \over 2}\cot {\pi \over {2\pi }}

Q39

In a triangle

ABC

, medians

AD

and

BE

are drawn. If

AD=4

,

\angle DAB = {\pi \over 6}

and

\angle ABE = {\pi \over 3}

, then the area of the

\angle \Delta ABC

is :

A

{{64} \over 3}

B

{8 \over 3}

C

{{16} \over 3}

D

{{32} \over {3\sqrt 3 }}

Correct Answer

Option D

Solution

AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,

Let

PB=x

\tan {60^ \circ } = {{8/3} \over x}

or

x = {8 \over {3\sqrt 3 }}

Area of

\Delta ABD

= {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}

$\therefore$ Area of

\Delta ABC

= 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}

\left[ \, \right.

As median of a

\Delta

divides it into two

\Delta 's

of equal area.

\left. \, \right]

Q40

If in a

\Delta ABC

a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},

then the sides

a, b

and

c

:

A satisfy

a+b=c

B are in A.P

C are in G.P

D are in H.P

Correct Answer

Option B

Solution

If

a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2}

a\left[ {\cos C + 1} \right] + c\left[ {\cos A + 1} \right] = 3b

\left( {a + c} \right) + \left( {a\cos C + c\cos \,B} \right) = 3b

a + c + b = 3b

or

a + c = 2b

or

a,b,c

are in

A.P.