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Properties of Triangle
Properties of Triangle
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Question

A straight line cuts off the intercepts OA=a\mathrm{OA}=\mathrm{a} and OB=b\mathrm{OB}=\mathrm{b} on the positive directions of xx-axis and yy axis respectively. If the perpendicular from origin OO to this line makes an angle of π6\frac{\pi}{6} with positive direction of yy-axis and the area of △OAB\triangle \mathrm{OAB} is 9833\frac{98}{3} \sqrt{3}, then a2−b2\mathrm{a}^{2}-\mathrm{b}^{2} is equal to :

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Solution

12ab=9833{1 \over 2}ab = {{98\sqrt 3 } \over 3} ⇒3ab=196 \Rightarrow \sqrt 3 ab = 196 ..... (i) OP=OBcos⁡30∘=OAcos⁡60∘OP = OB\cos 30^\circ = OA\cos 60^\circ ⇒b32=a2 \Rightarrow {{b\sqrt 3 } \over 2} = {a \over 2} ⇒3b=a \Rightarrow \sqrt 3 b = a ..... (ii) By (i) and (ii) a2=196{a^2} = 196 a=14a = 14 b2=a23{b^2} = {{{a^2}} \over 3} a2−b2=2a23=3923{a^2} - {b^2} = {{2{a^2}} \over 3} = {{392} \over 3}

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