Question

Triangle ABC shown below has mA = 92°, b = 10, and c = 14. Find the area of

the triangle.
Round your answer to the nearest tenth and do not include units in your answer.

Answer 1

70.0

Explanation:

To find the area of Triangle ABC, we can use the formula for the area of a triangle given two sides and the included angle. The formula is:

`\sf \textsf{Area} = (1)/(2) * b * c * \sin(A)`

where:

-
`\sf b` and
`\sf c` are the lengths of the two sides,

-
`\sf A` is the measure of the included angle.

In this case, we are given that
`\sf A = 92^\circ`,
`\sf b = 10`, and
`\sf c = 14`.

`\sf \textsf{Area} = (1)/(2) * 10 * 14 * \sin(92^\circ)`

Using a calculator:

`\sf \textsf{Area} \approx 70 * 0.999390827`

`\sf \textsf{Area} \approx 69.95735789`

`\sf \textsf{Area} \approx 70.0 \textsf{ ( in nearest tenth )}`

Therefore, the area of Triangle ABC is approximately 70.0 square units.

Answer 2

70.0

Explanation:

To find the area of a triangle given the lengths of two of its sides and the measure of the included angle, we can use the following formula:

`\large\boxed{\text{Area} = (1)/(2)\: b\:c \sin A}`

where:

• 
  `b` and
  `c` are the lengths of the two sides.
• 
  `A` is the measure of the included angle.

In the case of triangle ABC:

• b = 10
• c = 14
• A = 92°

Substitute the values into the formula:

`\begin{aligned}\text{Area\;$\triangle ABC$} &= (1)/(2) \cdot 10 \cdot 14\cdot \sin 92^(\circ)\\\\&=70\sin 92^(\circ)\\\\&=69.95735789....\\\\&=70.0\; \sf square\;units\;(nearest\;tenth)\end{aligned}`

Therefore, the area triangle ABC is 70.0 square units (rounded to the nearest tenth).