`DeltaABC` में, `angleBAC=90^(@)` तथा `AB=1/2BC` है, तब `angleACB` है।

Options

A. `60^(@)`
B. `30^(@)`
C. `45^(@)`
D. `15^(@)`

Answer

B

Explanation

According to question,
Given:
BAC is right angle triangle
`AB=1/2BC`

`(AB)/(BC)` = `1/2`
`sintheta =P/H=1/2`
`sin30^(@)=1/2`
`therefore theta=angleACB=30^(@)`