Question

solve each triangle. round your answers to the nearest tenth. 1) (mangle b = 25^{circ},mangle c = 127^{circ},b = 9) (mangle a=28^{circ},c = 17,a = 10) 3) (c = 18mathrm{cm},mangle a = 137^{circ},b = 12mathrm{cm}) (mangle b = 17^{circ},mangle c = 26^{circ},a = 28mathrm{cm})

Explanation:

Step1: Find angle $A$

In a triangle, the sum of interior angles is $180^{\circ}$. Given $m\angle B = 25^{\circ}$ and $m\angle C=127^{\circ}$, then $m\angle A=180^{\circ}-(25^{\circ} + 127^{\circ})=28^{\circ}$.

Step2: Use the Law of Sines to find side $a$

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}$. We know $b = 9$, $m\angle A = 28^{\circ}$, and $m\angle B=25^{\circ}$. So $a=\frac{b\sin A}{\sin B}=\frac{9\times\sin28^{\circ}}{\sin25^{\circ}}\approx\frac{9\times0.4695}{0.4226}\approx9.9$.

Step3: Use the Law of Sines to find side $c$

Using the Law of Sines $\frac{c}{\sin C}=\frac{b}{\sin B}$. We know $b = 9$, $m\angle B = 25^{\circ}$, and $m\angle C=127^{\circ}$. So $c=\frac{b\sin C}{\sin B}=\frac{9\times\sin127^{\circ}}{\sin25^{\circ}}\approx\frac{9\times0.7986}{0.4226}\approx16.9$.

Answer:

$m\angle A = 28^{\circ},a\approx9.9,c\approx16.9$