`DeltaABC` में, `angleC` अधिक कोण हे|कोण A और B के बाह्य समद्विभाजक , भुजा BC तथा AC को बिंदु D तथा E पर मिलते हे| यदि AB=CD हे, तब `angleACB=?`

Options

A. `105^(@)`
B. `108^(@)`
C. `110^(@)`
D. `135^(@)`

Answer

C

Explanation

According to question,

Let `angleCAB=x` and `angleCBA=y`
`rArr angleCAD=(180-x)/2 = 90-x/2`
and `angleEBC=(180-y)/2 = 90-y/2`
also `angleAEB=angleEAB=x`
`(therefore` AB=EB `rArr` ABE is an isosceles triangle)
and `angleADB = angleABD`=y(therefore AB=AD `rArr` ADB, is an isoceles triangle)
In `DeltaAEB`,
`angleAEB + angleABE + angleBAE = 180^(@)`
`x+x+y+90-y/2=180^(@)`
`rArr =4x+y=180^(@)`
Similarly in `DeltaADB`
`4x+y=180^(@)`
`rArr 4y+x+4x+y=180^(@)+180`
`rArr x+y=72^(@)`
In triangle ABC,
`rArr angleACB=180-72=108^(@)`