To solve the problem of finding the measure of [tex]\(\angle PQR\)[/tex] in [tex]\(\triangle PQR\)[/tex], given [tex]\(PQ = 7 \, \text{cm}\)[/tex], [tex]\(QR = 16 \, \text{cm}\)[/tex], and the area of [tex]\(\triangle PQR\)[/tex] is [tex]\(28 \sqrt{3} \, \text{cm}^2\)[/tex], follow these steps:
1. Formula for the area of a triangle using two sides and the included angle:
[tex]\[
\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR)
\][/tex]
2. Substitute the given values into the formula:
[tex]\[
28 \sqrt{3} = \frac{1}{2} \times 7 \, \text{cm} \times 16 \, \text{cm} \times \sin(\angle PQR)
\][/tex]
3. Solve for [tex]\(\sin(\angle PQR)\)[/tex]:
[tex]\[
28 \sqrt{3} = 56 \sin(\angle PQR)
\][/tex]
[tex]\[
\sin(\angle PQR) = \frac{28 \sqrt{3}}{56}
\][/tex]
[tex]\[
\sin(\angle PQR) = \frac{\sqrt{3}}{2}
\][/tex]
4. Find the angle whose sine is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]:
We know from trigonometric values that:
[tex]\[
\sin(60^\circ) = \frac{\sqrt{3}}{2}
\][/tex]
Therefore, [tex]\(\angle PQR = 60^\circ\)[/tex].
Thus, the measure of [tex]\(\angle PQR\)[/tex] is:
[tex]\[
\boxed{60^{\circ}}
\][/tex]
