Area of triangle inscribed in a circle

[Guest]

Let ABC be a triangle inscribed in a circle with radius 1 unit .
O is the centre and AD is the diameter . Both B and C are
located at the left side of AD . If AB is with length x unit
while AC with y unit and let x ≤ y . Find
(1) angle BAO
(2) angle CAO
(3) area of Δ ABC

[shyamjayakannan]

Screenshot 2026-03-13 200933.png (109.21 KiB) Viewed 64 times

Construct OB and OC as shown in the figure.

1) Now, apply the cosine rule in [tex]\triangle AOB[/tex]: [tex]\cos{\angle BAO}=\frac{{AO}^2+{AB}^2-{BO}^2}{2AO\times AB}\Rightarrow\cos{\angle BAO}=\frac{1+x^2-1}{2x}=\frac{x}{2}[/tex]
2) Similarly, [tex]\cos{\angle CAO}=\frac{y}{2}[/tex] doing the same in [tex]\triangle AOC[/tex]

Take the inverse cosine to get the angles. Now, for (3), [tex]ar(\triangle ABC)=\frac{1}{2}AB\times AC\sin{\angle BAC}=\frac{1}{2}xy\sin{(\angle BAO-\angle CAO)}[/tex]

[tex]=\frac{1}{2}xy(\sin{\angle BAO}\cos{\angle CAO}-\cos{\angle BAO}\sin{\angle CAO})=\frac{1}{2}xy\left\{\sqrt{1-\left(\frac{x}{2}\right)^2}\frac{y}{2}-\frac{x}{2}\sqrt{1-\left(\frac{y}{2}\right)^2}\right\}[/tex]

[tex]=\boxed{\frac{1}{8}xy\left(y\sqrt{4-x^2}-x\sqrt{4-y^2}\right)}[/tex]