Solve the right triangle.
-$$\mathrm { B } = 32.7 ^ { \circ } , \mathrm { c } = 4.6 \mathrm {~mm} , \mathrm { C } = 90 ^ { \circ }$$
Round values to one decimal place.
A) $\mathrm { a } = 3.9 \mathrm {~mm} , \mathrm {~A} = 57.3 ^ { \circ } , \mathrm { b } = 2.5 \mathrm {~mm}$
B) $\mathrm { a } = 3.1 \mathrm {~mm} , \mathrm {~A} = 57.3 ^ { \circ } , \mathrm { b } = 3.4 \mathrm {~mm}$
C) $\mathrm { a } = 3.9 \mathrm {~mm} , \mathrm {~A} = 57.3 ^ { \circ } , \mathrm { b } = 3.1 \mathrm {~mm}$
D) $\mathrm { a } = 2.5 \mathrm {~mm} , \mathrm {~A} = 57.3 ^ { \circ } , \mathrm { b } = 3.9 \mathrm {~mm}$

Question

Quizplus · Accepted Answer

Given $\mathrm { B } = 32.7 ^ { \circ }$ and $\mathrm { C } = 90 ^ { \circ }$, we can find $\mathrm { A } = 180 ^ { \circ } - 90 ^ { \circ } - 32.7 ^ { \circ } = 57.3 ^ { \circ }$. Using trigonometric ratios in a right triangle, $\sin(B) = \frac{b}{c}$ and $\sin(A) = \frac{a}{c}$, we can solve for sides $a$ and $b$. Given $c = 4.6 \mathrm {~mm}$, $\sin(32.7^\circ) = \frac{b}{4.6}$ gives $b \approx 2.5 \mathrm {~mm}$, and $\sin(57.3^\circ) = \frac{a}{4.6}$ gives $a \approx 3.9 \mathrm {~mm}$.

Solve the Right Triangle B=32.7∘,c=4.6 mm,C=90∘\mathrm { B } = 32.7 ^ { \circ } , \mathrm { c } = 4.6 \mathrm {~mm} , \mathrm { C } = 90 ^ { \circ }B=32.7∘,c=4.6 mm,C=90∘

Solve the Right Triangle $\mathrm { B } = 32.7 ^ { \circ } , \mathrm { c } = 4.6 \mathrm {~mm} , \mathrm { C } = 90 ^ { \circ }$