Problem of the Week Problem E and Solution The Angle Between

Problem

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.

Solution

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, $$\angle TQU \approx 26.6\degree$$.