`Delta ABC` का अतः केंद्र I है तथा `/_ABC = 90^(@)` और `/_ACB = 70^(@)` हो तो `/_BIC` कितना है?

Options

A. `115^(@)`
B. `100^(@)`
C. `110^(@)`
D. `105^(@)`

Answer

B

Explanation

`therefore` sum of all angles of a triangle =`180^(@)`
So, `angleBAC = 180-(90+70) = 20^(@)`

so, `angleBIC = 90+1/2angleA`
`=90^(@)+1/2 xx 20^(@)`
`angleBIC=100^(@)`