!!theta + beta = ?!!
We need to solve this question in two parts.
In the first figure of a circle we can find that the angle of the traingle at tangent point is !!90deg!! as diameter makes an angle !!90deg!! on the circle.
Now using straight line concepts we know sum of angle of a straight line is !!180deg!!
Hence, !!theta + 90 + 30 = 180!!
!!theta = 180 - 120!!
!!=> theta = 60!!
In the second part we can solve using tangent property of the circle.
!!BQ = BP!!, !!AP = AR!! and !!QC = CR!!
Also using pythagoras theorem we can get value of !!AC = sqrt(AB^2 + BC^2)!!
!!=> AC = sqrt(8^2 + 6^2)!!
!!=> AC = 10!!
Now Suppose !!AR = x!!
Hence !!AP = x!!, RC = (10-x) , QC = (10-x), BQ = (x-4) , PB = (x-4)!!
Hence, !!x + (x-4) = 8!!
!!=> 2x = 12!!
!!=> x = 6!!
Then, !!BQ = r = (x-4) = (6-4) = 2!!
It is given that !!r = cosecbeta!!
Therefore, !!cosecbeta = 2!!
!!=> sinbeta = 1/2!!
hence, !!beta = 30deg!!
So, !!theta + beta = 60deg + 30 deg = 90deg!!