# How do you solve the right triangle given A = 28° 30' , b = 18.3?

Oct 25, 2015

∠A= 28.5º, ∠B=61.5º, ∠C=90º
a≈4.2, b=18.3, c≈18.8

#### Explanation:

We are given this triangle:

Since there are 60 minutes in a degree, we know that ∠A= 28.5º

Since we know ∆ABC is a right triangle, ∠C=90º

Since there are 180º in all triangles, ∠A+∠B+∠C=180º, which means that 28.5º+∠B+90º=180º, which means that ∠B=61.5º

Now our triangle is:

Using the law of sines ($\sin \frac{A}{a} = \sin \frac{B}{b}$), we can obtain the length of side a: $\sin \frac{28.5}{a} = \sin \frac{61.5}{18.3}$

This simplifies to sin(61.5)×a=18.3sin(28.5), which simplifies to $a = \frac{18.3 \sin \left(28.5\right)}{\sin} \left(61.5\right)$, which is approximately 4.2.

Now our triangle looks like this:

From here, we can solve for side c by using the pythagorean theorem (${a}^{2} + {b}^{2} = {c}^{2}$): ${\left(18.3\right)}^{2} + {\left(4.2\right)}^{2} = {c}^{2}$

c=sqrt((18.3)^2+(4.2)^2)=sqrt(352.53)≈18.8

So we end with our solved triangle that looks like this: