InterviewSolution
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InterviewSolution
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Reduce the equation `sqrt3+y+2=0` to the normal form `x cos alpha+y sin alpha=p`, and hence find the value of `alpha and p`. |
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Answer» We have `sqrt3x+y+2=0 Rightarrow -sqrt3x-y=2 " "["keeping constant +ve"]` `Rightarrow ((-sqrt3)/(2))x+((-1)/(2))y=1 ` `["on dividing throughout by"sqrt((sqrt3)^(2)+1^(2))]` `Rightarrow x cos alpha+y sin alpha=p,"where "cos alpha=(-1)/(2), sin alpha=(-sqrt3)/(2) and p=(5)/(2)` Since, `cos alpha lt 0 and sin alpha lt 0, alpha lies in third quadrant. Now, `tan alpha=(sin alpha)/(cos alpha)=((-sqrt3)/(2))xx(-2)=sqrt3=tan(180^(@)+60^(@))=tan 240^(@)` Thus, ``alpha=240^(@) and p=(5)/(2)` Hence, the given equation in normal form is `x cos 240^(@)+y sin 240^(@)=(5)/(2)` |
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