Properties of Triangle — Page 7 | JEE Mathematics

Q61

Let the area of a

\triangle P Q R

with vertices

P(5,4), Q(-2,4)

and

R(a, b)

be 35 square units. If its orthocenter and centroid are

O\left(2, \dfrac{14}{5}\right)

and

C(c, d)

respectively, then

c+2 d

is equal to

A

3

B

\dfrac{7}{3}

C

2

D

\dfrac{8}{3}

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Equation of lines } Q R=5 x+2 y+2=0 \\ & \text{ Equation of lines } P R=10 x-3 y-38=0 \\ & \therefore \text{ Point } R(2,-6) \\ & \text{ Centroid }=\left(\frac{5-2+2}{3}, \frac{4+4-6}{3}\right) \\ & =\left(\frac{5}{3}, \frac{2}{3}\right) \\ & c+2 d=\frac{5}{3}+\frac{4}{3}=3 \end{aligned}

Q62

A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. Fromthe top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feetof the towers T1 and T2 is :

A

10\sqrt 2

B

10\sqrt 3

C

20\sqrt 3

D

20\sqrt 2

Correct Answer

Option C

Solution

Let the distance between T1 and T2 be x From the figure EA = 60 m (T1) and DB = 80 m (T2)

\angle DEC = \theta

and

\angle BEC = 2\theta

Now in

\angle DEC

,

\tan \theta = {{DC} \over {AB}} = {{20} \over x}

and in

\Delta BEC

,

\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}

We know that

\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}

$\Rightarrow$

{{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}

$\Rightarrow$

{x^2} = 1200

$\Rightarrow$

x = 20\sqrt 3

Q63

The angle of elevation of the top of a vertical tower from a point A, due east of it is 45o. The angle of elevation of the top of the same tower from a point B, due south of A is 30o. If the distance between A and B is

54\sqrt 2 \,m,

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

A

36\sqrt 3

B 54

C

54\sqrt 3

D 108

Correct Answer

Option B

Solution

Let the height of tower = h In triangle PQA, tan45o =

{h \over {QA}}

$\Rightarrow$ h = QA In triangle PQB, tan300 =

{h \over {BQ}}

$\Rightarrow$ BQ =

\sqrt 3

h In triangle BAQ $\angle$ QAB = 90o. $\therefore$ QA2 + AB2 = QB2 $\Rightarrow$ h2 + (54

\sqrt 2

)2 = (

\sqrt 3

h)2 $\Rightarrow$ 2h2 = (54

\sqrt 2

)2 $\Rightarrow$

\sqrt 2 h

= 54

\sqrt 2

$\Rightarrow$ h = 54 m

Q64

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30o to 45o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

A

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

B

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

C

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

D

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

Correct Answer

Option A

Solution

Assume height of tower = AB = h From

\Delta

ABD, tan45o =

{{AB} \over {AD}}

$\Rightarrow$

\,\,\,

{{AB} \over {AD}} = 1

[ as tan45o = 1] $\Rightarrow$

\,\,\,

AB = AD

\therefore\,\,\,

AD = h From

\Delta

BAC, tan30o =

{{AB} \over {AC}}

$\Rightarrow$

\,\,\,

{h \over {AC}} = {1 \over {\sqrt 3 }}

$\Rightarrow$

\,\,\,

AC = h

\sqrt 3

As, AC = AD + DC $\Rightarrow$

\,\,\,

DC = AC $-$ AD =

\sqrt 3 h

$-$ h Given that, time taken to reach from point C to D = 18 min.

\therefore\,\,\,

Car speed =

{{distance} \over {time}}

=

{{CD} \over {18}}

=

{{(\sqrt 3 - 1)h} \over {18}}

\therefore\,\,\,

Time taken to move from D to A =

{{Distan ce\,\,of\,\,DA} \over {speed}}

=

{h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}}

=

{{18} \over {\left( {\sqrt 3 - 1} \right)}}

=

{{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}}

= 9

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

min.

Q65

Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30o at B. The height (in m) of the lamp-post is -

A

2\sqrt {21}

B

{3 \over 2}\sqrt {21}

C

7\sqrt {3}

D

{2 \over 3}\sqrt {21}

Correct Answer

Option D

Solution

BD = hcot30o = h

\sqrt 3

So, 72 + 52 = 2(h

\sqrt 3

)2 + 32) $\Rightarrow$ 37 = 3h2 + 9 $\Rightarrow$ 3h2 = 28 $\Rightarrow$ h =

\sqrt {{{28} \over 3}} = {2 \over 3}\sqrt {21}

Q66

If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30o and the angle of depression of reflection of the cloud in the lake from P be 60o, then the height of the cloud (in meters) from the surface of the lake is :

A 45

B 42

C 50

D 60

Correct Answer

Option C

Solution

tan 30o =

{x \over y} \Rightarrow y = \sqrt 3 x\,\,\,\,....(i)

tan 60o =

{{25 + x + 25} \over y}

\Rightarrow \,\,\sqrt 3 y = 50 + x

$\Rightarrow$

3x = 50 + x

$\Rightarrow$ x = 25 m $\therefore$ Height of cloud from surface = 25 + 25 = 50m

Q67

In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 – c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :

A

{y \over {\sqrt 3 }}

B

{c \over 3}

C

{c \over {\sqrt 3 }}

D

{3 \over 2}

y

Correct Answer

Option C

Solution

Given a + b = x and ab = y If x2 $-$ c2 = y $\Rightarrow$ (a + b)2 $-$ c2 = ab $\Rightarrow$ a2 + b2 $-$ c2 = $-$ ab $\Rightarrow$

{{{a^2} + {b^2} - {c^2}} \over {2ab}} = - {1 \over 2}

\Rightarrow \cos C = - {1 \over 2}

\Rightarrow \angle C = {{2\pi } \over 3}

R = {c \over {2\sin C}} = {c \over {\sqrt 3 }}

Q68

Given

{{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}

for a

\Delta

ABC with usual notation. If

{{\cos A} \over \alpha } = {{\cos B} \over \beta } = {{\cos C} \over \gamma },

then the ordered triad (

\alpha

,

\beta

,

\gamma

) has a value :

A (19, 7, 25)

B (7, 19, 25)

C (5, 12, 13)

D (3, 4, 5)

Correct Answer

Option B

Solution

b + c = 11 $\lambda$ , c + a = 12 $\lambda$ , a + b = 13 $\lambda$ $\Rightarrow$ a = 7 $\lambda$ , b = 6 $\lambda$ , c = 5 $\lambda$ (using cosine formula) cosA =

{1 \over 5},

cosB =

{19 \over 35},

cosC =

{5 \over 7},

$\alpha$ : $\beta$ : $\gamma$ $\Rightarrow$ 7 : 19 : 25