Question

b
a
30°
c
a
c
b=10
∠b=30°
b=10
a=?
c=?
∠a=?

Explanation:

Step1: Find ∠A

In a right - triangle, the sum of the interior angles is \(180^{\circ}\), and one angle is \(90^{\circ}\), \(\angle B = 30^{\circ}\). So \(\angle A=180^{\circ}-\ 90^{\circ}-30^{\circ}=60^{\circ}\)

Step2: Find side \(a\) (adjacent to \(30^{\circ}\) angle)

In a \(30^{\circ}-60^{\circ}-90^{\circ}\) right - triangle, \(\tan B=\frac{b}{a}\) (since \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), here for \(\angle B\), opposite side is \(b\) and adjacent side is \(a\)). We know that \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\) and \(b = 10\). So \(\frac{1}{\sqrt{3}}=\frac{10}{a}\), then \(a = 10\sqrt{3}\)

Step3: Find side \(c\) (hypotenuse)

In a \(30^{\circ}-60^{\circ}-90^{\circ}\) right - triangle, the hypotenuse \(c = 2b\) (because the side opposite \(30^{\circ}\) is half of the hypotenuse, here \(b\) is opposite \(30^{\circ}\) angle? Wait, no. Wait, in the right - triangle, the right angle is at \(C\), so the sides: \(b\) is the side opposite \(\angle A\)? Wait, no, let's re - define. Let's use trigonometric ratios correctly. \(\cos B=\frac{a}{c}\) and \(\sin B=\frac{b}{c}\). Since \(\sin30^{\circ}=\frac{1}{2}\) and \(b = 10\), from \(\sin B=\frac{b}{c}\), we have \(\frac{1}{2}=\frac{10}{c}\), so \(c = 20\) (or we can use the Pythagorean theorem: \(c^{2}=a^{2}+b^{2}\), \(a = 10\sqrt{3}\), \(b = 10\), then \(c^{2}=(10\sqrt{3})^{2}+10^{2}=300 + 100=400\), so \(c = 20\))

Answer:

\(\angle A = 60^{\circ}\), \(a = 10\sqrt{3}\), \(c = 20\)