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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
1: Join O and B
2: Join O and C
Consider side of square as 'a'
As shown in figure below
As we can see, OB is diagonal of square as well as radius of semicircle.
⇒ r = a√2
And also angle OFC is 135
°
Now apply cosine rule on Triangle OFC
\[cos(135)=\frac{a^{2}+CF^{2}-(a\sqrt{2})^2}{2a\times CF}\]
\[-\frac{1}{\sqrt{2}}=\frac{a^{2}+CF^{2}-2a^2}{2a\times CF}\]
\[-1=\frac{CF^{2}-a^2}{a\sqrt{2}\times CF}\]
\[-a\sqrt{2} CF=CF^{2}-a^2\]
\[CF^{2}+a\sqrt{2} CF-a^2=0\]
Now apply Brahmagupta's Quadratic Formula
\[CF=\frac{-\sqrt{2}a\pm\sqrt{2a^2+4a^2}}{2}\]
\[CF=\frac{-\sqrt{2}a\pm\sqrt{6}a}{2}\]
\[CF=\frac{-a\pm\sqrt{3}a}{\sqrt{2}}\]
CF must be positive
\[\Rightarrow CF=\frac{\sqrt{3}a-a}{\sqrt{2}}\]
\[\Rightarrow CF=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)a\]
Now apply Sine rule in triangle OFC
\[\frac{a\sqrt{2}}{sin(135)}=\frac{CF}{sin(y)}\]
\[\frac{a\sqrt{2}}{\frac{1}{\sqrt{2}}}=\frac{\frac{\sqrt{3}-1}{\sqrt{2}}a}{sin(y)}\]
\[2a\times sin(y)=\frac{\sqrt{3}-1}{\sqrt{2}}a\]
\[sin(y)=\frac{\sqrt{3}-1}{2\sqrt{2}}\]
\[sin(y)=15^{o}\]
Now observe Triangle COD
CO = OD -------(Both are radius of circle)
Angle COD = 90- Angle COF = 90-15
Angle COD = 75°
Triangle COD Is Isosceles Triangle
⇒ Angle OCD = Angle ODC = 52.5
⇒ q°= 52.5
135 + 15 + p = 180
⇒ p° = 30
Angle x° = p°+q° = 30+52.5
x° = 82.5°
