Question

determine the number of triangles possible in each case with the given information.
a) \\( \angle a = 30 ^ { \circ }, a = 4 \mathrm { cm }, b = 8 \mathrm { cm } \\)
b) \\( \angle a = 14 ^ { \circ }, a = 15 \mathrm { cm }, b = 24 \mathrm { cm } \\)
c) \\( \angle a = 143 ^ { \circ }, a = 7 \mathrm { cm }, b = 8 \mathrm { cm } \\)
d) \\( \angle a = 45 ^ { \circ }, a = 26 \mathrm { cm }, b = 13 \mathrm { cm } \\)

Explanation:

Step1: Use the sine law formula

The sine law formula is \(\frac{a}{\sin A}=\frac{b}{\sin B}\). For each case, we can find \(\sin B\) and then determine the number of possible triangles.

Case a)

Given \(A = 30^{\circ}\), \(a = 4\mathrm{cm}\), \(b = 8\mathrm{cm}\)
Using the sine law \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we substitute the values:
\(\sin B=\frac{b\sin A}{a}\)
\(\sin B=\frac{8\times\sin30^{\circ}}{4}\)
Since \(\sin30^{\circ}=\frac{1}{2}\), we have \(\sin B=\frac{8\times\frac{1}{2}}{4}= 1\)
Since \(\sin B = 1\), \(B = 90^{\circ}\), so there is \(1\) triangle.

Case b)

Given \(A=14^{\circ}\), \(a = 15\mathrm{cm}\), \(b = 24\mathrm{cm}\)
Using the sine law \(\sin B=\frac{b\sin A}{a}\)
\(\sin B=\frac{24\times\sin14^{\circ}}{15}\)
\(\sin14^{\circ}\approx0.2419\)
\(\sin B=\frac{24\times0.2419}{15}\approx0.387\)
\(B_1=\sin^{- 1}(0.387)\approx22.8^{\circ}\)
\(B_2 = 180^{\circ}-22.8^{\circ}=157.2^{\circ}\)
\(A + B_2=14^{\circ}+157.2^{\circ}=171.2^{\circ}<180^{\circ}\)
So there are \(2\) triangles.

Case c)

Given \(A = 143^{\circ}\), \(a = 7\mathrm{cm}\), \(b = 8\mathrm{cm}\)
Using the sine law \(\sin B=\frac{b\sin A}{a}\)
\(\sin A=\sin143^{\circ}\approx0.6018\)
\(\sin B=\frac{8\times0.6018}{7}\approx0.688\)
\(B_1=\sin^{-1}(0.688)\approx43.5^{\circ}\)
\(A + B_1=143^{\circ}+43.5^{\circ}=186.5^{\circ}>180^{\circ}\)
So there are \(0\) triangles.

Case d)

Given \(A = 45^{\circ}\), \(a = 26\mathrm{cm}\), \(b = 13\mathrm{cm}\)
Using the sine law \(\sin B=\frac{b\sin A}{a}\)
\(\sin A=\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071\)
\(\sin B=\frac{13\times0.7071}{26}\approx0.3536\)
\(B_1=\sin^{-1}(0.3536)\approx20.7^{\circ}\)
\(B_2=180^{\circ}-20.7^{\circ}=159.3^{\circ}\)
\(A + B_2=45^{\circ}+159.3^{\circ}=204.3^{\circ}>180^{\circ}\)
So there is \(1\) triangle.

Answer:

a) \(1\)
b) \(2\)
c) \(0\)
d) \(1\)