Math / Geometry

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Find the perimeter of $\triangle P Q R$ . Round your answer to the nearest tenth if necessary. Figures are necessarily drawn to scale.

Studdy Solution

STEP 1

1. We are given a triangle $\triangle PQR$ .
2. The side lengths $PQ = 23.4$ , $PR = 27$ , and $QR = x$ (unknown).
3. The angles are $\angle P = 49^\circ$ , $\angle Q = 74^\circ$ , and $\angle R = 57^\circ$ .
4. We need to find the perimeter of the triangle, which is the sum of its side lengths.

STEP 2

1. Verify the given information and check if the angles add up to $180^\circ$ .
2. Use the Law of Sines to find the unknown side $QR$ .
3. Calculate the perimeter of the triangle by summing all side lengths.

STEP 3

Verify the angles of the triangle. The sum of the angles in a triangle should be $180^\circ$ :
$\angle P + \angle Q + \angle R = 49^\circ + 74^\circ + 57^\circ = 180^\circ$
The angles add up correctly, confirming the triangle's validity.

STEP 4

Use the Law of Sines to find the unknown side $QR$ . The Law of Sines states:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
For $\triangle PQR$ , we have:
$\frac{QR}{\sin P} = \frac{PR}{\sin Q}$
Substitute the known values:
$\frac{x}{\sin 49^\circ} = \frac{27}{\sin 74^\circ}$

STEP 5

Solve for $x$ (side $QR$ ):
$x = \frac{27 \cdot \sin 49^\circ}{\sin 74^\circ}$
Calculate the value using a calculator:
$x \approx \frac{27 \cdot 0.7547}{0.9613}$ $x \approx \frac{20.3769}{0.9613}$ $x \approx 21.2$

STEP 6

Calculate the perimeter of $\triangle PQR$ by summing the side lengths:
$\text{Perimeter} = PQ + PR + QR$ $= 23.4 + 27 + 21.2$ $= 71.6$
The perimeter of $\triangle PQR$ is approximately:
$\boxed{71.6}$

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