Problem E and Solution

The Angle Between

Rectangle \(PQRS\) has \(PQ=3\) and \(QR=4\). Points \(T\) and \(U\) are on side \(PS\) such that \(PT=US=1\).

Determine the measure of \(\angle TQU\), in degrees and rounded to \(1\) decimal place.

Let \(X\) be the point on \(QR\) such that \(UX\) is parallel to \(SR\). Then \(\angle UXQ = 90\degree\). Also, \(UX=SR=3\), \(XR=US=1\), and therefore \(QX=3\).

It follows that \(\triangle UXQ\) is an isosceles right-angled triangle, and so \(\angle UQX=\angle QUX=45\degree\).

From here we present three different solutions.

**Solution 1**

Since \(\triangle TPQ\) is a right-angled triangle, \(\tan(\angle TQP)=\frac{1}{3}\), and so \(\angle TQP \approx 18.4\degree\).

Since \(\angle PQX=90\degree\), we can calculate the value of \(\angle TQU\) as follows. $$\begin{align} \angle PQX &= \angle PQT + \angle TQU + \angle UQX\\ \angle TQU &= \angle PQX - \angle PQT - \angle UQX\\ \angle TQU &\approx 90\degree - 18.4\degree - 45\degree\\ \angle TQU &\approx 26.6\degree \end{align}$$ Therefore, \(\angle TQU \approx 26.6\degree\).

**Solution 2**

Since \(\triangle TPQ\) is a right-angled triangle, by the Pythagorean Theorem, \(TQ^2=PT^2+PQ^2=1^2+3^2=10\). Therefore \(TQ=\sqrt{10}\), since \(TQ>0\).

Since \(PQRS\) is a rectangle, \(PS=QR=4\), and \(PT=US=1\), it follows that \(TU=2\). Since \(\triangle UXQ\) is a right-angled triangle, by the Pythagorean Theorem, \(QU^2=UX^2+QX^2=3^2+3^2=18\). Therefore \(QU=\sqrt{18}\), since \(QU>0\).

Now we will use the cosine law in \(\triangle TQU\).

$$\begin{align} TU^2&=TQ^2+QU^2-2(TQ)(QU)\cos(\angle TQU)\\ 2^2&=10+18-2(\sqrt{10})(\sqrt{18})\cos(\angle TQU)\\ 4-10-18&=-2\sqrt{10}\sqrt{18}\cos(\angle TQU)\\ -24&=-2\sqrt{10}\sqrt{18}\cos(\angle TQU)\\ \frac{12}{\sqrt{10}\sqrt{18}}&=\cos(\angle TQU)\\ \angle TQU&\approx 26.6\degree \end{align}$$

Therefore, \(\angle TQU \approx 26.6\degree\).

**Solution 3**

Since \(\triangle TPQ\) is a right-angled triangle, by the Pythagorean Theorem, \(TQ^2=PT^2+PQ^2=1^2+3^2=10\). Therefore \(TQ=\sqrt{10}\), since \(TQ>0\).

Since \(PQRS\) is a rectangle, \(PS=QR=4\), and \(PT=US=1\), it follows that \(TU=2\).

Since \(UX\) is parallel to \(SR\), then \(\angle PUX=90\degree\). Since \(\angle QUX=45\degree\), it follows that \(\angle TUQ=45\degree\).

Now we will use the sine law in \(\triangle TQU\). $$\begin{align} \frac{\sin(\angle TQU)}{TU}&=\frac{\sin(\angle TUQ)}{TQ}\\ \frac{\sin(\angle TQU)}{2}&=\frac{\sin 45\degree}{\sqrt{10}}\\ \angle TQU &\approx 26.6\degree \end{align}$$ Therefore, \(\angle TQU \approx 26.6\degree\).