Question

1. find the length of bc to the nearest tenth of a centimetre.

Explanation:

Step1: Find BD in right - triangle ABD

In right - triangle ABD, $\sin A=\frac{BD}{AB}$. Given $A = 26^{\circ}$ and $AB = 22.9$ cm. So, $BD=AB\sin A$. Substituting the values, we have $BD = 22.9\times\sin26^{\circ}$. Since $\sin26^{\circ}\approx0.4384$, then $BD=22.9\times0.4384\approx10.04$ cm.

Step2: Find BC in right - triangle BCD

In right - triangle BCD, $\sin C=\frac{BD}{BC}$. Given $C = 49^{\circ}$ and we found $BD\approx10.04$ cm. Then $BC=\frac{BD}{\sin C}$. Substituting $BD\approx10.04$ cm and $\sin C=\sin49^{\circ}\approx0.7547$, we get $BC=\frac{10.04}{0.7547}\approx13.3$ cm.

Answer:

$13.3$ cm