Question

calculate the value of bd correct to 2 decimal places.

Explanation:

Step1: Find BC in right - triangle ABC

In right - triangle ABC, $\sin\angle A=\frac{BC}{AB}$. Given $AB = 8$ and $\angle A=25^{\circ}$, so $BC = AB\sin\angle A=8\times\sin25^{\circ}$.
Since $\sin25^{\circ}\approx0.4226$, then $BC = 8\times0.4226 = 3.3808$.

Step2: Find BD in right - triangle BCD

In right - triangle BCD, $\sin\angle D=\frac{BC}{BD}$. Given $\angle D = 15^{\circ}$ and $BC = 3.3808$. Then $BD=\frac{BC}{\sin\angle D}=\frac{3.3808}{\sin15^{\circ}}$.
Since $\sin15^{\circ}=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588$.
So $BD=\frac{3.3808}{0.2588}\approx13.06$.

Answer:

$13.06$