Question

if the midpoints of the longest side of the triangle formed by the line 4x+3y= 12 , x=0 , y=0 is ( 3sinalpha, 4cosBita), then cot(3alpha-bita) is​

Answer 1

Step-by-step explanation:

Hello dear my name is Ravi Gupta from Ayodhya Uttar Pradesh India

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Your Solution is Here

Draw the line on Cartesian plane

X- axis showing equation

y=0

Y- axis showing equation

x=0

and drawing equation 4x+3y=12

We changes it in intercepts form of equation

we get

x/3+y/4= 1

here we get the point where line cut X-axis and Y-axis also

the point where line cut X-axis (3,0)

and Y-axis (0,4)

then mid point is=(3/2,2)

but mid point is given=( 3sin@,cosBita)

here we evaluate Alpha and bita by equating above express

Alpha=30°,bita=60°

then we put vlaue of Alpha and bita in question

$3 \sin( 3\alpha - \beta$

we get answer=√3

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