FIND THE ACUTE ANGLE BETWEEN THE FOLLOWING PAIRS OF LINE (2+√3y)x+y-5=0 and (2+√3)x-y+3=0 chapter name is STRAIGHT LINES FROM STD XI WE HAVE TO FIND TAN θ EXPLAIN ME BRIEFLY WITH EACH STEP

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You need to find the acute angle theta between the given lines using the following formula, such that:

`tan theta = |(m_1 - m_2)/(1 + m_1*m_2)|`

You need to find the slopes `m_1` and `m_2` , hence, you need to convert the given formas of the lines into slope intercept form, such that:

`(2+sqrt3)x+y-5=0 => y = -(2+sqrt3)x + 5 => m_1 = -(2+sqrt3)`

`(2+sqrt3)x-y+3=0 => y = (2+sqrt3)x + 3 => m_2 = (2+sqrt3)`

`tan theta = |(-(2+sqrt3) - (2+sqrt3))/(1 - (2+sqrt3)^2)|`

`tan theta = |(-2(2+sqrt3))/(1 - 2 - 4sqrt3 - 3)|`

`tan theta = |(-2(2+sqrt3))/(-4 - 4sqrt3)|`

`tan theta = |(-2(2 + sqrt3))/(-2(2 + 2sqrt3))|`

`tan theta = |(2 + sqrt3)/(2 + 2sqrt3)|`

`theta = arctan |(4 - 4sqrt3 + 2sqrt3 - 6)/(-8)|`

`theta = arctan |(-2 - 2sqrt3)/(-8)|`

`theta = arctan |(1 + sqrt3)/4| => theta ~~ 34^o`

Hence, evaluating the acute angle between the given lines yields `theta ~~ 34^o.`

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