Question

HEY GUYS⭐⭐
PLZZZ HELP IN MY Q'S
CLASS 11
ITS VERY URGENT
I'LL MARK AS BRAINLIEST ☺️☺️NO SPAMMING ⭕⭕
EXPLAIN PROPERLY⏩⏩

Answer 1

Hope u like my process
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Straight line Equations are in form of
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=> a x + by + c = 0

Given:
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=> (cos²A)x + (cos A)y + 1 =0 ____(1)

=> (cos²B)x + (cosB)y + 1 =0 _____(2)

=> (cos²C)x + (cosC)y + 1 = 0 _____(3)

where (1), (2),(3) are concurrent.
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For concurrency we know that
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Determinant of the a, b, c of all 3 straight lines will be 0.

For isosceles triangle..
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Any of the two angles of triangle should be equal.
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Here determinant is taken out by elementary row operations where I first did R1` = R1 - R2 and R2` =R2 - R3

Plz.. Follow the pic...

Sorry I though to explain here by writings but time expired.
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Hope this is ur required answer

Proud to help you ☺️

Answer 2

Hey friend, Harish here.

Here is your answer.

Given that the three following lines are concurrent :

$\big( \cos^2 \mathrm A \big )x + \big (\cos \mathrm A \big )y+1 = 0 \\ \\ \big( \cos^2 \mathrm B \big )x + \big (\cos \mathrm B \big )y+1 = 0 \\ \\ \big( \cos^2 \mathrm C \big )x + \big (\cos \mathrm C \big )y+1 = 0$

So as they are concurrent we can represent the following determinant.

$\begin{vmatrix}\cos^2A&\cos A&1\\\cos^2B&\cos B&1\\\cos^2C&\cos C&1\\\end{vmatrix}=0.$

And it is interesting to note that it is a 3 x 3 Vandermonde determinant which equals :

$(\cos\mathrm A-\cos\mathrm B)(\cos\mathrm A-\cos\mathrm C)(\cos\mathrm B-\cos\mathrm C) = 0$

Hence atleast two of the cosines or equal.

So, As two angles are atleast equal the triangle formed with these angles must be a isosceles triangle.

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Hope my answer is helpful to you.  

Thanks @RakeshMohata for you answer. Just simplified into few steps for clear understanding.