In figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and `angleBQR=70^(@)` then `angleAQB` is equal to

Options

A. `20^(@)`
B. `40^(@)`
C. `35^(@)`
D. `45^(@)`

Answer

B

Explanation

Given, AB||PR

`:.angleAQB =angleBQR=70^(@)` [alternate angles]
Also, QD is perpendicular to AB and QD bisects AB.
In `DeltaQDA and DeltaQDB`, `angleQDA=angleQDB` [each `90^(@)`]
AD=BD
QD=QD [common side]
`:.DeltaADQ=DeltaBDQ` [by SAS similarity criterion]
Then `angleQAD=angleQBD` [CPCT.....(i)]
Also `angleABQ=angleBQR` [altermat interior angle]
`:.angleABQ=70^(@)` `[:' angleBQR=70^(@)]`
Hence, `angleQAB=70^(@)` [from Eq. (i)]
Now, in `DeltaABQ`, `angleA+angleB+angleQ=180^(@)`
`rArrangleQ=180^(@)-(70^(@)+70^(@))=40^(@)`