Help with problem

[green island]

We have an isosceles triangle inscribed in a circle (that is, all its vertices lie on the circumference of the circle).
The vertex angle of the triangle — the angle between the two equal sides — is \theta.
The circle has radius R=1.

Express the area of the triangle as a function of \theta.

(Hint: You may use the formula for the area of a triangle inscribed in a circle: \tfrac{abc}{4R}, where a, b, c are the side lengths. But this formula is not mandatory. The result should simplify to the form \sin\theta(1+\cos\theta)

[pastel thorn]

We have an isosceles triangle inscribed in a circle (that is, all its vertices lie on the circumference of the circle).\
The vertex angle of the triangle — the angle between the two equal sides — is $\theta$.\
The circle has radius R=1.\

Express the area of the triangle as a function of \theta.\

(Hint: You may use the formula for the area of a triangle inscribed in a circle: $\frac{abc}{4R}$, where a, b, c are the side lengths. But this formula is not mandatory. The result should simplify to the form $\sin\theta (1+\cos\theta)$

[tiny marten]

This is sad 😢

We have an isosceles triangle inscribed in a circle (that is, all its vertices lie on the circumference of the circle).\\
The vertex angle of the triangle — the angle between the two equal sides — is $\theta$.\\
The circle has radius R=1.\\

Express the area of the triangle as a function of \theta.\\

(Hint: You may use the formula for the area of a triangle inscribed in a circle: $\frac{abc}{4R}$, where a, b, c are the side lengths. But this formula is not mandatory. The result should simplify to the form $\sin\theta (1+\cos\theta)$
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l.53 ...ea of the triangle as a function of \theta
                                                  .\\
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[past onyx]

start by drawing the 3 radius of the circle by connecting the centre to each of the 3 vertices of the isosceles triangle

[green island]

Ok so now we have 3 triangles with every triangle having 2 sides equal to the radius so = 1

[past onyx]

right