# Derive formula for the area of triangle provided that you know 2 angles and side between them.

I am trying to derive the formula for finding the area of the triangle that uses 2 angles and and 1 side between them.

Formula provided by the book:

If we know two angles (call them $$A$$ and $$B$$) and the side between them call it $$b$$, then Area = $$\frac{b^2 \sin B \sin C}{2 \sin A}$$

However, I keep getting different result.

My attempt as follows:

First we draw following picture:

$$\angle YPX = 90^{\circ}$$ and $$\angle YPZ = 90^{\circ}$$

We know that $$\angle YXZ = A$$ and $$\angle YZX = C$$ and XZ = $$b$$

Let $$XP = d$$ and $$PZ = b-d$$

$$\tan A = \frac{YP}{XP} = \frac{YP}{d} \implies YP = \tan A \cdot d$$

$$\tan C = \frac{YP}{PZ} = \frac{YP}{b-d} \implies YP = \tan C \cdot (b-d)$$

It follows that $$\tan A \cdot d = \tan C \cdot (b-d)$$ $$d = \frac{\tan C \cdot (b-d)}{\tan A}$$ $$\frac{d}{b-d} = \frac{\tan C}{\tan A}$$ $$\frac{b-d}{d} = \frac{\tan A}{\tan C}$$ $$\frac{b}{d} - 1 = \frac{\tan A}{\tan C}$$ $$\frac{b}{d} = \frac{\tan A}{\tan C} + 1$$ $$\frac{1}{d} = \frac{\tan A}{\tan C \cdot b} + \frac{1}{b}$$ $$\frac{1}{d} = \frac{\tan A}{\tan C \cdot b} + \frac{\tan C}{\tan C \cdot b}$$ $$\frac{1}{d} = \frac{\tan A + \tan C}{\tan C \cdot b}$$ $$d= \frac{\tan C \cdot b}{\tan A + \tan C}$$ $$d= \frac{\frac{\sin C}{\cos C} \cdot b}{\frac{\sin A}{\cos A} + \frac{\sin C}{\cos C}}$$ $$d= \frac{\frac{\sin C}{\cos C} \cdot b}{\frac{\sin A \cos C + \sin C \cos A}{\cos C \cos A} }$$

$$d = \frac{\frac{\sin C}{\cos C} \cdot b \cdot \cos C \cos A}{\sin A \cos C + \sin C \cos A}$$ $$d = \frac{\sin C \cos A\cdot b}{\sin A \cos C + \sin C \cos A}$$

We know that $$YP = \tan A \cdot d$$ $$YP = \frac{\sin A}{\cos A} \frac{\sin C \cos A\cdot b}{\sin A \cos C + \sin C \cos A}$$ $$YP = \frac{\sin A}{\cos A} \frac{\sin C \cos A\cdot b}{\sin(A+C)}$$ $$YP = \frac{\sin A \sin C \cdot b}{\sin(A+C)}$$

Because $$\text{Area} =\frac{1}{2} \cdot YP \cdot b$$

Then $$\text{Area} =\frac{1}{2} \cdot \frac{\sin A \sin C \cdot b}{\sin(A+C)} \cdot b$$ $$\text{Area} = \frac{\sin A \sin C \cdot b^2}{2 \sin(A+C)}$$

Which, as can be seen, is different from the one provided by the textbook. I've tried to find areas of several triangles using formula above and this calculator, both give the same results. However, I'm still unsure whether it is correct. Thus I would like to ask you, is the the derivation (and the formula) above correct?

## 1 Answer

It is simply use are of triangle is ∆=(1/2)cbSinA=(1/2)abSinB Therefore ∆^2 = (1/4) acb^2 SinA.SinB Also ∆=(1/2)acSinC Dividing both we get Area ∆ =( c^2 SinA.SinB)/2 Sin C