Question

find the exact value of each part labeled with a variable in the figure

Explanation:

Step1: Analyze the isosceles right triangle (left)

In the left triangle with angles \(45^\circ\), \(45^\circ\), \(90^\circ\), it's an isosceles right triangle, so \(p = 21\) (legs are equal). Also, use the Pythagorean theorem or trigonometric ratios to find \(r\). Using \(\sin(45^\circ)=\frac{21}{r}\), so \(r=\frac{21}{\sin(45^\circ)}=\frac{21}{\frac{\sqrt{2}}{2}} = 21\sqrt{2}\).

Step2: Analyze the right triangle with \(30^\circ\) angle (right)

In the right triangle with angle \(30^\circ\) and hypotenuse \(r = 21\sqrt{2}\), and angle \(45^\circ\) at the right angle vertex? Wait, no, the right triangle has a \(30^\circ\) angle, right angle, and the other angle is \(60^\circ\)? Wait, no, looking at the figure, the right triangle (with \(q\)) has hypotenuse \(r = 21\sqrt{2}\) and angle \(30^\circ\) at the top. So \(\sin(30^\circ)=\frac{q}{r}\), so \(q = r\sin(30^\circ)=21\sqrt{2}\times\frac{1}{2}=\frac{21\sqrt{2}}{2}\)? Wait, no, maybe I misread. Wait, the left triangle: angles \(45^\circ\), \(45^\circ\), \(90^\circ\), so legs are equal, so \(p = 21\). Then \(r\) is the hypotenuse of that triangle, so \(r = 21\sqrt{2}\) (since in isosceles right triangle, hypotenuse \(= leg\times\sqrt{2}\)). Then the right triangle: angle \(30^\circ\), hypotenuse \(r = 21\sqrt{2}\), and \(q\) is opposite \(30^\circ\)? Wait, no, the right triangle has a right angle, \(30^\circ\), so the side opposite \(30^\circ\) is half the hypotenuse? Wait, no, the hypotenuse of the right triangle (with \(q\)) is \(r\)? Wait, maybe the right triangle has angle \(30^\circ\), and the side adjacent to \(30^\circ\) is \(q\)? Wait, no, let's re - express:

Left triangle: right - angled, angles \(45^\circ\), \(45^\circ\), \(90^\circ\), one leg is \(21\), so the other leg \(p = 21\) (isosceles right triangle), and hypotenuse \(r=\sqrt{21^{2}+21^{2}}=\sqrt{2\times21^{2}} = 21\sqrt{2}\).

Right triangle: right - angled, angle \(30^\circ\), hypotenuse \(r = 21\sqrt{2}\), and the side \(q\) is opposite the \(30^\circ\) angle? Wait, no, in a right triangle, \(\sin(30^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}\), so if the angle is \(30^\circ\), and hypotenuse is \(r = 21\sqrt{2}\), then the side opposite \(30^\circ\) is \(q=\frac{1}{2}r=\frac{21\sqrt{2}}{2}\)? Wait, no, maybe the angle is \(30^\circ\), and the side adjacent to \(30^\circ\) is \(q\)? Wait, no, let's check the angles again. The top angle is \(30^\circ\), the right triangle has a right angle, so the other angle is \(60^\circ\). Wait, maybe I made a mistake. Let's start over.

Left triangle: angles \(45^\circ\), \(45^\circ\), \(90^\circ\), leg length \(21\). So:

- For \(p\): In isosceles right triangle, legs are equal, so \(p = 21\).

- For \(r\): Using Pythagorean theorem \(r=\sqrt{21^{2}+21^{2}}=\sqrt{441 + 441}=\sqrt{882}=21\sqrt{2}\) (since \(882=441\times2\), \(\sqrt{441\times2}=21\sqrt{2}\)).

Right triangle: hypotenuse \(r = 21\sqrt{2}\), angle \(30^\circ\) at the top, right angle at the bottom - right. So the side \(q\) is opposite the \(30^\circ\) angle. In a right triangle, \(\sin(30^\circ)=\frac{q}{r}\), so \(q = r\times\sin(30^\circ)=21\sqrt{2}\times\frac{1}{2}=\frac{21\sqrt{2}}{2}\). Wait, but also, we can check the other side, but the problem asks for \(p\), \(q\), \(r\). Wait, maybe the right triangle has angle \(30^\circ\), and the side \(q\) is opposite \(30^\circ\), and the hypotenuse is \(r\).

Wait, another approach: The left triangle is isosceles right - angled, so \(p = 21\), \(r = 21\sqrt{2}\). Then in the right triangle, which is a \(30 - 60 - 90\) triangle, with hypotenuse \(r = 21\sqrt{2}\), the side opposite \(30^\circ\) (which is \(q\)) is half the hypotenuse? Wait, no, in a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^\circ\) is the shortest side (length \(x\)), opposite \(60^\circ\) is \(x\sqrt{3}\), and hypotenuse is \(2x\). So if hypotenuse is \(r = 21\sqrt{2}\), then the side opposite \(30^\circ\) (which is \(q\)) is \(x=\frac{r}{2}=\frac{21\sqrt{2}}{2}\), and the side opposite \(60^\circ\) would be \(x\sqrt{3}=\frac{21\sqrt{6}}{2}\), but we don't need that.

So:

- \(p = 21\) (since it's an isosceles right triangle, legs are equal)

- \(r = 21\sqrt{2}\) (hypotenuse of isosceles right triangle, \(leg\times\sqrt{2}\))

- \(q=\frac{21\sqrt{2}}{2}\) (side opposite \(30^\circ\) in \(30 - 60 - 90\) triangle, \(\frac{1}{2}\times\) hypotenuse)

Answer:

\(p = 21\), \(r = 21\sqrt{2}\), \(q=\frac{21\sqrt{2}}{2}\)