Solve the right triangle.
-$$\mathrm { A } = 15 ^ { \circ } 19 ^ { \prime } , \mathrm { c } = 295 \mathrm { ft } , \mathrm { C } = 90 ^ { \circ }$$
Round side lengths to two decimal places, if necessary.
A) $\mathrm { B } = 74 ^ { \circ } 40 ^ { \prime } , \mathrm { a } = 81.13 \mathrm { ft } , \mathrm { b } = 285.72 \mathrm { ft }$
B) $\mathrm { B } = 74 ^ { \circ } 41 ^ { \prime } , \mathrm { a } = 77.36 \mathrm { ft } , \mathrm { b } = 283.63 \mathrm { ft }$
C) $B = 74 ^ { \circ } 41 ^ { \prime } , \mathrm { a } = 77.93 \mathrm { ft } , \mathrm { b } = 284.52 \mathrm { ft }$
D) $B = 75 ^ { \circ } 40 ^ { \prime } , \mathrm { a } = 77.93 \mathrm { ft } , \mathrm { b } = 280.52 \mathrm { ft }$

Question

Quizplus · Accepted Answer

Given $A = 15^\circ 19'$ and $C = 90^\circ$, angle $B = 180^\circ - A - C = 74^\circ 41'$. Using the sine rule or Pythagoras' theorem, we can calculate the sides $a$ and $b$, where $a = c \cdot \sin(A)$ and $b = c \cdot \cos(A)$, leading to the values close to those in option C.

Solve the Right Triangle A=15∘19′,c=295ft,C=90∘\mathrm { A } = 15 ^ { \circ } 19 ^ { \prime } , \mathrm { c } = 295 \mathrm { ft } , \mathrm { C } = 90 ^ { \circ }A=15∘19′,c=295ft,C=90∘

Solve the Right Triangle $\mathrm { A } = 15 ^ { \circ } 19 ^ { \prime } , \mathrm { c } = 295 \mathrm { ft } , \mathrm { C } = 90 ^ { \circ }$