Calculation of triangles side B if I know the angles and side a?

Question

How do I calculate the side B of the triangle if I know the following:

Side $A = 15 \rm {cm} ;\beta = 12^{\circ} ;\gamma= 90^{\circ} ;\alpha = 78^{\circ} $

Thank you.

Answer 1

For a $\triangle\text{ABC}$, we know that (for EVERY triangle):

$$ \begin{cases} \angle\alpha^\circ+\angle\beta^\circ+\angle\gamma^\circ=180^\circ\\ \\ \frac{\left|\text{A}\right|}{\sin\angle\alpha}=\frac{\left|\text{B}\right|}{\sin\angle\beta}=\frac{\left|\text{C}\right|}{\sin\angle\gamma}\\ \\ \left|\text{A}\right|^2=\left|\text{B}\right|^2+\left|\text{C}\right|^2-2\left|\text{B}\right|\left|\text{C}\right|\cos\angle\alpha\\ \\ \left|\text{B}\right|^2=\left|\text{A}\right|^2+\left|\text{C}\right|^2-2\left|\text{A}\right|\left|\text{C}\right|\cos\angle\beta\\ \\ \left|\text{C}\right|^2=\left|\text{A}\right|^2+\left|\text{B}\right|^2-2\left|\text{A}\right|\left|\text{B}\right|\cos\angle\gamma \end{cases}\tag1 $$

So, knowing your values:

$$ \begin{cases} 78^\circ+12^\circ+90^\circ=180^\circ\\ \\ \frac{15}{\sin\left(78^\circ\right)}=\frac{\left|\text{B}\right|}{\sin\left(12^\circ\right)}=\frac{\left|\text{C}\right|}{\sin\left(90^\circ\right)}\\ \\ 15^2=\left|\text{B}\right|^2+\left|\text{C}\right|^2-2\left|\text{B}\right|\left|\text{C}\right|\cos\left(78^\circ\right)\\ \\ \left|\text{B}\right|^2=15^2+\left|\text{C}\right|^2-2\cdot15\cdot\left|\text{C}\right|\cos\left(12^\circ\right)\\ \\ \left|\text{C}\right|^2=15^2+\left|\text{B}\right|^2-2\cdot15\cdot\left|\text{B}\right|\cos\left(90^\circ\right) \end{cases}\tag2 $$

So:

• $$\left|\text{B}\right|=\frac{15\cdot\sin\left(12^\circ\right)}{\sin\left(78^\circ\right)}\tag3$$
• $$\left|\text{C}\right|=\frac{15\cdot\sin\left(90^\circ\right)}{\sin\left(78^\circ\right)}=15\cdot\left\{2-\sqrt{5}+\sqrt{3\cdot\left(5-2\sqrt{5}\right)}\right\}\tag4$$

Answer 2

use that $$\frac{\sin(\beta)}{\sin(\alpha)}=\frac{b}{a}$$ and so $$b=\frac{a\sin(\beta)}{\sin(\alpha)}$$