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RD Chapter 11 Co ordinate Geometry Ex 11.1 Solutions

Question - 11 : - In a тИЖABC, тИаABC = тИаACB and the bisectors of тИаABC and тИаACB intersect at O such that тИаBOC = 120┬░. Show that тИаA = тИаB = тИаC = 60┬░.

Answer - 11 : -

Given : In тИаABC, BO and CO are the bisectors of тИаB and тИаC respectively and тИаBOC = 120┬░ and тИаABC = тИаACB
┬а
To prove : тИаA = тИаB = тИаC = 60┬░
Proof : тИ╡ BO and CO are the bisectors of тИаB and тИаC
тИ┤ тИаBOC = 90┬░ +┬а┬атИаA
But тИаBOC = 120┬░
тИ┤ 90┬░+┬а ┬атИаA = 120┬░
тИ┤ 12 тИаA = 120┬░ тАУ 90┬░ = 30┬░
тИ┤ тИаA = 60┬░
тИ╡ тИаA + тИаB + тИаC = 180┬░ (Angles of a triangle)
тИаB + тИаC = 180┬░ тАУ 60┬░ = 120┬░ and тИаB = тИаC
тИ╡ тИаB = тИаC =┬а┬а= 60┬░
Hence тИаA = тИаB = тИаC = 60┬░

Question - 12 : - If each angle of a triangle is less than the sum ofthe other two, show that the triangle is acute angled.

Answer - 12 : -

In a тИЖABC,
Let тИаA < тИаB + тИаC
┬а
тЗТтИаA + тИаA < тИаA + тИаB + тИаC
тЗТ 2тИаA < 180┬░
тЗТ тИаA < 90┬░ (тИ╡ Sum of angles of a triangle is 180┬░)
Similarly, we can prove that
тИаB < 90┬░ and тИаC < 90┬░
тИ┤ Each angle of the triangle are acute angle.

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