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The area of a triangle is 3 . Two of the side lengths are 1.8 and 4.8 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.
Answer:

The area of a triangle is $3$ . Two of the side lengths are $1$ . $8$ and $4$ . $8$ and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree. $\newline$ Answer:

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Find missing angles in special triangles

Full solution

Q. The area of a triangle is

3

. Two of the side lengths are

1

.

8

and

4

.

8

and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.

\newline

Answer:

Area Formula: To find the measure of the included angle in a triangle given its area and two side lengths, we can use the formula for the area of a triangle, which is $A = \frac{1}{2}ab\sin(C)$ , where $A$ is the area, $a$ and $b$ are the side lengths, and $C$ is the included angle.
Calculate Product: We know the area $A = 3$ , side length $a = 1.8$ , and side length $b = 4.8$ . We can plug these values into the area formula to solve for $\sin(C)$ . $\newline$ $3 = \frac{1}{2} \times 1.8 \times 4.8 \times \sin(C)$
Divide Area: First, calculate the product of $\frac{1}{2}$ , $1.8$ , and $4.8$ . $\newline$ $\frac{1}{2} \times 1.8 \times 4.8 = 4.32$
Find Sin(C): Now, divide the area by this product to solve for $\sin(C)$ . $\newline$ $\sin(C) = \frac{3}{4.32}$
Inverse Sine: Perform the division to find $\sin(C)$ . $\newline$ $\sin(C) = 0.6944...$
Calculate Angle: To find the angle $C$ , we need to take the inverse sine (arcsin) of $\sin(C)$ . $\newline$ $C = \arcsin(0.6944...)$
Calculate Angle: To find the angle $C$ , we need to take the inverse sine (arcsin) of $\sin(C)$ . $\newline$ $C = \arcsin(0.6944...)$ Using a calculator, we find the value of $C$ to the nearest tenth of a degree. $\newline$ $C \approx 43.9^\circ$

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