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1) tan(5π/6) = tan (π -π/6) = -tan (π/6) = - 1/√3
2) (3-5)/(l-3) = 2
=> -2/(l-3) = 2
=> l-3 = -1,
=> l= 2
3) slope of line joining (-3,2) and (5, -4) is (-4-2)/(5+3) = -6/8
hence, slope of line perpendicular to it will be 8/6 = 4/3
4) slope of line is tan of acute angle made by that line with x axis, so, let required angle be ©, then, tan (©) = √3,
=> © = 60° or π/3
2) (3-5)/(l-3) = 2
=> -2/(l-3) = 2
=> l-3 = -1,
=> l= 2
3) slope of line joining (-3,2) and (5, -4) is (-4-2)/(5+3) = -6/8
hence, slope of line perpendicular to it will be 8/6 = 4/3
4) slope of line is tan of acute angle made by that line with x axis, so, let required angle be ©, then, tan (©) = √3,
=> © = 60° or π/3
MrMk:
phone me root wese hi bnta h
sorry I didn't see negative sign
hn, bhai wo uper h,
OK
and in second question, i replaced lambda by l,
please check 4 th one
the answer is option b
answer is incorrect, I'm 100% sure,
you can confirm with your teacher
ok
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