5. Let $A B C$ and $P Q R$ be two triangles. If $\cos A=\sin P, \cos B=\sin Q$ and $\cos C=\sin R$, what is the largest angle (in degrees) among the six interior angles of the two triangles?
(1 mark)
設 $A B C$ 和 $P Q R$ 為三角形。若 $\cos A=\sin P 、 \cos B=\sin Q$ 且 $\cos C=\sin R$,則兩個三角形六個內角中最大的一個（以「度」為單位）是多少？
5. $135^{\circ}$
5. Since $\sin P, \sin Q$ and $\sin R$ are all positive, so are $\cos A, \cos B$ and $\cos C$. In other words, $\triangle A B C$ is acute-angled. It follows that we must have $P=90^{\circ} \pm A, Q=90^{\circ} \pm B$ and $R=90^{\circ} \pm C$. Since $P Q R$ is a triangle, at most one of $P, Q, R$ can be obtuse. If all of $P, Q, R$ are acute, then $180^{\circ}=P+Q+R=\left(90^{\circ}-A\right)+\left(90^{\circ}-B\right)+\left(90^{\circ}-C\right)=270^{\circ}-180^{\circ}=90^{\circ}$, which is a contradiction. We conclude that exactly one of $P, Q, R$ is obtuse. So without loss of generality we assume that $P=90^{\circ}+A, Q=90^{\circ}-B$ and $R=90^{\circ}-C$. In this case, we must have $180^{\circ}=P+Q+R=\left(90^{\circ}+A\right)+\left(90^{\circ}-B\right)+\left(90^{\circ}-C\right)=270^{\circ}+A-\left(180^{\circ}-A\right)=90^{\circ}+2 A$. It follows that $A=45^{\circ}$ and so the largest angle is $P=135^{\circ}$.