Question

the following figure shows △abc with side lengths to the nearest tenth. find m∠a. note that m∠a is acute. round to the nearest degree. m∠a = \boxed{\space}°

Explanation:

Step1: Identify Law of Sines

In $\triangle ABC$, we use the Law of Sines: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Here, side $BC = a = 8$, side $AC = b$ (not given), side $AB = c = 15$, $\angle B = 28^\circ$, $\angle A$ is what we need to find. So $\frac{BC}{\sin A}=\frac{AB}{\sin B}$.

Step2: Substitute values

Substitute $BC = 8$, $AB = 15$, $\angle B = 28^\circ$ into the formula: $\frac{8}{\sin A}=\frac{15}{\sin 28^\circ}$.

Step3: Solve for $\sin A$

Cross - multiply: $15\sin A=8\sin 28^\circ$. Then $\sin A=\frac{8\sin 28^\circ}{15}$. Calculate $\sin 28^\circ\approx0.4695$, so $8\sin 28^\circ\approx8\times0.4695 = 3.756$. Then $\sin A=\frac{3.756}{15}\approx0.2504$.

Step4: Find $\angle A$

Since $\angle A$ is acute, $A=\arcsin(0.2504)\approx14.5^\circ\approx15^\circ$ (rounded to the nearest degree).

Answer:

$15$