Question

calculate the value of bd correct to 2 decimal places.

Explanation:

Step1: Find BC in right - triangle ABC

In right - triangle ABC, $\sin A=\frac{BC}{AB}$. Given $AB = 12$ and $A = 15^{\circ}$, so $BC=AB\sin A=12\sin15^{\circ}$.
Since $\sin15^{\circ}=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx\frac{2.449 - 1.414}{4}=\frac{1.035}{4}=0.25875$, then $BC = 12\times0.25875 = 3.105$.

Step2: Find BD in right - triangle BCD

In right - triangle BCD, $\sin D=\frac{BC}{BD}$. Given $D = 18^{\circ}$ and $BC = 3.105$, so $BD=\frac{BC}{\sin D}$.
Since $\sin18^{\circ}\approx0.309$, then $BD=\frac{3.105}{0.309}\approx10.05$.

Answer:

$10.05$