Question

find the missing parts of the triangle.

a. b = 90°, c = 60°, b = 9√3
b. no such triangle exists.
c. b = 60°, c = 90°, b = 9
d. b = 60°, c = 90°, b = 9√3

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, \(a = 9\), \(A = 30^{\circ}\), \(c = 18\).
Substitute into \(\frac{a}{\sin A}=\frac{c}{\sin C}\):
\(\frac{9}{\sin30^{\circ}}=\frac{18}{\sin C}\)
Since \(\sin30^{\circ}=\frac{1}{2}\), we have \(\frac{9}{\frac{1}{2}}=\frac{18}{\sin C}\), which simplifies to \(18=\frac{18}{\sin C}\), so \(\sin C = 1\), and \(C = 90^{\circ}\).

Step2: Find angle \(B\)

Since the sum of angles in a triangle is \(180^{\circ}\), \(A + B + C=180^{\circ}\).
Substitute \(A = 30^{\circ}\) and \(C = 90^{\circ}\): \(30^{\circ}+B + 90^{\circ}=180^{\circ}\), so \(B = 60^{\circ}\).

Step3: Find side \(b\)

Using the Law of Sines again \(\frac{b}{\sin B}=\frac{a}{\sin A}\).
\(\sin B=\sin60^{\circ}=\frac{\sqrt{3}}{2}\), \(\sin A=\sin30^{\circ}=\frac{1}{2}\), \(a = 9\).
\(\frac{b}{\frac{\sqrt{3}}{2}}=\frac{9}{\frac{1}{2}}\), cross - multiply: \(b\times\frac{1}{2}=9\times\frac{\sqrt{3}}{2}\), so \(b = 9\sqrt{3}\).

Answer:

D. \(B = 60^{\circ},C = 90^{\circ},b = 9\sqrt{3}\)