`DeltaABC` के `angleB` का आतंरिक तथा `angleC` का बाह्य समद्विभाजक बिंदु D पर प्रतिछेदित होता है। यदि `angleBDC=50^(@)` है, तब `angleA` ?

Options

A. `100^(@)`
B. `90^(@)`
C. `120^(@)`
D. `60^(@)`

Answer

A

Explanation

According to question,

`angleBAD= 109^(@)`
`angleACB= 72^(@)`
`therefore angleACD = 180^(@)-72^(@)`
`angleACD=108^(@)`
`therefore AC=CD`
`angleCAD=angleCDA`
In angleCDA
`angleCAD+ angleCDA + angleDCA = 180^(@)`
`2angleCAD+108^(@)=180^(@)`
`2angleCAD=180^(@)-108^(@)`
`2angleCAD=72^(@)`
`angleCAD=72^(@)/2`
`angleCAD=36^(@)`
`therefore angleCAB=109^(@)-36^(@)`
`angleCAB=73^(@)`
In `DeltaABC`
`angleABC + angleACB + angleCAB= 180^(@)`
`angleABC=180^(@)-145^(@)`
`angleCAB=73^(@)`