Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l }
\mathrm { b } = 9.1 \mathrm { ft } \
\mathrm { A } = 35 ^ { \circ } 20 ^ { \prime } \
\mathrm { C } = 106 ^ { \circ } 20 ^ { \prime }
\end{array}$$
A) $37 \mathrm { ft } ^ { 2 }$
B) $79 \mathrm { ft } ^ { 2 }$
C) $32 \mathrm { ft } ^ { 2 }$
D) $74 \mathrm { ft } ^ { 2 }$

Question

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
\mathrm { b } = 9.1 \mathrm { ft } \
\mathrm { A } = 35 ^ { \circ } 20 ^ { \prime } \
\mathrm { C } = 106 ^ { \circ } 20 ^ { \prime }
\end{array}$$
A) $37 \mathrm { ft } ^ { 2 }$
B) $79 \mathrm { ft } ^ { 2 }$
C) $32 \mathrm { ft } ^ { 2 }$
D) $74 \mathrm { ft } ^ { 2 }$

Quizplus · Accepted Answer

The area of a triangle is given by the formula $A = \frac{1}{2}bh$, where $b$ is the length of the base and $h$ is the height. In this case, the base is $9.1 \mathrm { ft }$ and the height is the length of the altitude from vertex $A$ to side $BC$. To find the height, we can use the sine function:$$\sin A = \frac{h}{c}$$where $c$ is the length of the side opposite angle $A$. In this case, $c = 106 ^ { \circ } 20 ^ { \prime }$ and $\sin A = \sin 35 ^ { \circ } 20 ^ { \prime }$. Solving for $h$, we get:$$h = c \sin A = 106 ^ { \circ } 20 ^ { \prime } \sin 35 ^ { \circ } 20 ^ { \prime } \approx 58.1 \mathrm { ft }$$Therefore, the area of the triangle is:$$A = \frac{1}{2}bh = \frac{1}{2}(9.1 \mathrm { ft })(58.1 \mathrm { ft }) \approx 37 \mathrm { ft } ^ { 2 }$$So the correct answer is A.

Find the Area of Triangle ABC with the Given Parts