Here's how to solve the problem: Problem 6: Calculate $\phi$ and $\theta$, and then find the size of $\angle PQR$ to the nearest degree. Solution: The diagram shows a triangle $PQR$ with an altitude from $Q$ to $PR$. Let the foot of the altitude be $M$. We have two right-angled triangles: $\triangle PQM$ and $\triangle RQM$. Given: $PM = 9$ $MR = 23$ $QM = 10$ $\angle PQM = \theta$ $\angle RQM = \phi$ $\angle PQR = \theta + \phi$ Step 1: Calculate $\theta$ using $\triangle PQM$. In $\triangle PQM$, the side opposite to $\theta$ is $PM = 9$, and the side adjacent to $\theta$ is $QM = 10$. We use the tangent function: $$ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{PM}{QM} $$ $$ \tan(\theta) = \frac{9}{10} $$ $$ \theta = \arctan\left(\frac{9}{10}\right) $$ $$ \theta \approx 41.987^\circ $$ Step 2: Calculate $\phi$ using $\triangle RQM$. In $\triangle RQM$, the side opposite to $\phi$ is $MR = 23$, and the side adjacent to $\phi$ is $QM = 10$. We use the tangent function: $$ \tan(\phi) = \frac{\text{opposite}}{\text{adjacent}} = \frac{MR}{QM} $$ $$ \tan(\phi) = \frac{23}{10} $$ $$ \phi = \arctan\left(\frac{23}{10}\right) $$ $$ \phi \approx 66.501^\circ $$ Step 3: Calculate $\angle PQR$. The angle $\angle PQR$ is the sum of $\theta$ and $\phi$: $$ \angle PQR = \theta + \phi $$ $$ \angle PQR \approx 41.987^\circ + 66.501^\circ $$ $$ \angle PQR \approx 108.488^\circ $$ Step 4: Round $\angle PQR$ to the nearest degree. $$ \angle PQR \approx 108^\circ $$ The values are: $\boxed{\theta \approx 42.0^\circ}$ (to one decimal place) $\boxed{\phi \approx 66.5^\circ}$ (to one decimal place) The size of $\angle PQR$ to the nearest degree is $\boxed{108^\circ}$.