Question

the following figure shows △abc with side lengths to the nearest tenth. find ( mangle a ). note that ( mangle a ) is acute. round to the nearest degree. ( mangle a = square^circ ) (figure shows △abc with ( bc = 15 ), ( ac = 8 ), ( angle b = 25^circ ))

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 $BC = 15$ (opposite $\angle A$), side $AC = 8$ (opposite $\angle B = 25^\circ$). So $\frac{BC}{\sin A}=\frac{AC}{\sin B}$.

Step2: Substitute values

Substitute $BC = 15$, $AC = 8$, $\angle B = 25^\circ$ into the formula: $\frac{15}{\sin A}=\frac{8}{\sin 25^\circ}$.

Step3: Solve for $\sin A$

Cross - multiply: $8\sin A=15\sin 25^\circ$. Then $\sin A=\frac{15\sin 25^\circ}{8}$. Calculate $\sin 25^\circ\approx0.4226$, so $15\times0.4226 = 6.339$, and $\sin A=\frac{6.339}{8}\approx0.7924$.

Step4: Find $\angle A$

Since $\angle A$ is acute, $A=\arcsin(0.7924)\approx52^\circ$.

Answer:

$52$