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question

The vertex angle of an isosceles triangle is $\theta$. The sides of the triangle are $\sin\theta,\sqrt{\sin\theta}$, and $\sqrt{\sin\theta}$. Give the numeric value of the area of the triangle.

I've set up the law of cosines but that's all i have currently, any suggestions?

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    The law of cosines doesn't have anything to do with the area directly. The law of sines, on the other hand, is directly derived from a certain formula for the area of a triangle. Do you know which one? – 2017-01-25
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    @Did That's what happen if you (read I) don't read the whole question;) – 2017-01-25
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    @John Rawls why don't you react to the question of Arthur ? Interaction is encouraged on this site... – 2017-01-25

2 Answers 2

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We may suppose that $$AB=AC=\sqrt{\sin\theta},\quad BC=\sin\theta,\quad \angle{BAC}=\theta$$

Then, considering $\triangle{ABD}$ where $D$ is the midpoint of the side $BC$ of $\triangle{ABC}$, we have $$\sin\frac{\theta}{2}=\frac{\frac{\sin\theta}{2}}{\sqrt{\sin\theta}},$$ i.e. $$2\sin\frac{\theta}{2}=\sqrt{\sin\theta}$$ Squaring the both sides gives $$4\sin^2\frac{\theta}{2}=\sin\theta,$$ i.e. $$2(1-\cos\theta)=\sin\theta$$ Squaring the both sides gives $$4(1-\cos\theta)^2=1-\cos^2\theta,$$ i.e. $$(5\cos\theta-3)(\cos\theta-1)=0$$ So, having $\cos\theta=\frac{3}{5}$ gives that the area is $$\frac{1}{2}\sqrt{\sin\theta}\sqrt{\sin\theta}\ \sin\theta=\frac{\sin^2\theta}{2}=\frac{1-\cos^2\theta}{2}=\color{red}{\frac{8}{25}}$$

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    I don't get the last line. How do you figure the area to be $\sqrt{\sin\theta}\sqrt{\sin\theta}\sin \theta$? The base being $\sin\theta$? But then the height should be $\sqrt{\sin\theta}\cos(\theta/2)$ – 2017-01-25
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    @skyking: The area is $\color{red}{\frac 12}\sqrt{\sin\theta}\sqrt{\sin\theta}\ \sin\theta$, not $\sqrt{\sin\theta}\sqrt{\sin\theta}\ \sin\theta$. This comes from that $\frac 12 AB\times AC\sin\angle{BAC}$. See [here](https://en.wikipedia.org/wiki/Triangle#Using_trigonometry) for the formula. – 2017-01-25
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    +1. The fact that $\cos \theta = 3/5$ makes me wonder if there's a clever way to show why the $3$-$4$-$5$ triangle should be involved in this problem. – 2017-01-25
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Let's call $\angle BAC=\theta$, $BC=\sin \theta$ and $R$ the radius of the circumcircle. Using Sine Rule we have:

$$\frac{BC}{\sin \theta}=2R \to R=\frac{1}{2}$$

So the area is:

$$\text{Area}=\frac{\sin \theta \cdot \sqrt{\sin \theta}\cdot\sqrt{\sin \theta}}{4R}=\frac{\sin^2 \theta}{2}$$

Using Cossine rule:

$$\sin^2 \theta=\sin \theta+\sin \theta-2\sin \theta \cos \theta \to 2\cos \theta=2-\sin \theta \\\to 4\cos^2 \theta=4-4\sin \theta+\sin^2 \theta \to \sin \theta=\frac{4}{5}$$

so,

$$\text{Area}=\frac{8}{25}$$