Question

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

Explanation:

Step1: Analyze the isosceles right triangle

The left triangle is a right - isosceles triangle (one angle is \(90^{\circ}\), another is \(45^{\circ}\), so the third is also \(45^{\circ}\)). In an isosceles right triangle, the legs are equal. So \(p = 21\). Also, for an isosceles right triangle with leg length \(a\), the hypotenuse \(r\) is given by \(r=a\sqrt{2}\). Substituting \(a = 21\), we get \(r = 21\sqrt{2}\).

Step2: Analyze the 30 - 60 - 90 triangle

Now, consider the right triangle with angle \(30^{\circ}\) and hypotenuse \(r = 21\sqrt{2}\). In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the side opposite the \(30^{\circ}\) angle (which is \(q\)) is half of the hypotenuse. So \(q=\frac{1}{2}r\). Substituting \(r = 21\sqrt{2}\), we get \(q=\frac{21\sqrt{2}}{2}\)? Wait, no, wait. Wait, the angle at the top is \(30^{\circ}\), the right - angled triangle with hypotenuse \(r\) and one angle \(30^{\circ}\): the side opposite \(30^{\circ}\) is \(q\), and the side adjacent to \(30^{\circ}\) is the other leg. Wait, actually, in the right triangle with hypotenuse \(r\) and angle \(30^{\circ}\), \(\sin(30^{\circ})=\frac{q}{r}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(q = r\times\sin(30^{\circ})\). But we know from the isosceles right triangle that \(r = 21\sqrt{2}\)? Wait, no, wait. Wait, the left triangle: angles are \(90^{\circ}\), \(45^{\circ}\), \(45^{\circ}\), so it's an isosceles right triangle, so legs are equal, so \(p = 21\), and hypotenuse \(r\): by Pythagoras, \(r=\sqrt{21^{2}+21^{2}}=\sqrt{2\times21^{2}} = 21\sqrt{2}\). Then the other right triangle: angle at the top is \(30^{\circ}\), right angle, so the angle opposite \(q\) is \(30^{\circ}\), so \(\sin(30^{\circ})=\frac{q}{r}\), so \(q = r\times\sin(30^{\circ})=21\sqrt{2}\times\frac{1}{2}=\frac{21\sqrt{2}}{2}\)? Wait, no, maybe I mixed up the angles. Wait, the triangle with hypotenuse \(r\) and angle \(30^{\circ}\): let's re - examine the figure. The left triangle has angles \(90^{\circ}\), \(45^{\circ}\), \(45^{\circ}\), so the side with length 21 and \(p\) are the legs, so \(p = 21\), \(r\) is the hypotenuse of this isosceles right triangle, so \(r = 21\sqrt{2}\). Then the other triangle: it's a right triangle with hypotenuse \(r\) and one angle \(30^{\circ}\), and the side \(q\) is opposite the \(30^{\circ}\) angle. So in a \(30 - 60 - 90\) triangle, the side opposite \(30^{\circ}\) is half the hypotenuse. So \(q=\frac{r}{2}=\frac{21\sqrt{2}}{2}\)? Wait, no, that can't be. Wait, maybe the left triangle: the angle of \(45^{\circ}\) is at the top, so the triangle with sides 21, \(p\), and \(r\) is an isosceles right triangle, so \(p = 21\), \(r = 21\sqrt{2}\). Then the other triangle: angle at the top is \(30^{\circ}\), so the angle between \(r\) and the vertical side is \(30^{\circ}\), and the right - angled triangle with \(q\) as the leg opposite \(30^{\circ}\), so \(q=r\sin(30^{\circ})=21\sqrt{2}\times\frac{1}{2}=\frac{21\sqrt{2}}{2}\), and the other leg (adjacent to \(30^{\circ}\)) would be \(r\cos(30^{\circ})\), but we don't need that. Wait, but maybe I made a mistake. Wait, let's start over.

First, the left triangle: right - angled, one angle \(45^{\circ}\), so it's an isosceles right triangle. So the two legs are equal. So if one leg is 21, the other leg \(p = 21\). The hypotenuse \(r\) of an isosceles right triangle with leg length \(a\) is \(r=a\sqrt{2}\), so \(r = 21\sqrt{2}\).

Now, the right - angled triangle with hypotenuse \(r\) and angle \(30^{\circ}\): in a right - angled triangle, \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The side \(q\) is opposite the \(30^{\circ}\) angle, so \(\sin(30^{\circ})=\frac{q}{r}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(q=\frac{1}{2}r\). Substituting \(r = 21\sqrt{2}\), we get \(q=\frac{21\sqrt{2}}{2}\)? Wait, no, that seems odd. Wait, maybe the angle is \(60^{\circ}\)? Wait, no, the problem says the angle at the top is \(30^{\circ}\). Wait, maybe the left triangle is not the isosceles right triangle. Wait, the left triangle has angles \(90^{\circ}\), \(45^{\circ}\), so the third angle is \(45^{\circ}\), so it is isosceles. So legs are equal, so \(p = 21\), hypotenuse \(r = 21\sqrt{2}\). Then the other triangle: angle \(30^{\circ}\), right angle, so the side \(q\) is opposite \(30^{\circ}\), so \(q=\frac{r}{2}=\frac{21\sqrt{2}}{2}\), and \(r = 21\sqrt{2}\), \(p = 21\).

Wait, let's verify with trigonometry. For the isosceles right triangle: \(\cos(45^{\circ})=\frac{21}{r}\), so \(r=\frac{21}{\cos(45^{\circ})}=\frac{21}{\frac{\sqrt{2}}{2}}=21\sqrt{2}\), which matches. And \(\tan(45^{\circ})=\frac{p}{21}\), since \(\tan(45^{\circ}) = 1\), so \(p = 21\).

For the other triangle: \(\sin(30^{\circ})=\frac{q}{r}\), so \(q=r\sin(30^{\circ})=21\sqrt{2}\times\frac{1}{2}=\frac{21\sqrt{2}}{2}\).

Answer:

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