Question

8)
in the diagram, we have triangle rpq with angle at r being 35°, angle at q being 95°, and we need to find the angle at p between pg and pq (marked as ?).

Explanation:

Step1: Recall exterior angle theorem

The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.

Step2: Identify angles for the theorem

In triangle \(RQP\), the angle at \(Q\) is \(95^{\circ}\), the angle at \(R\) is \(35^{\circ}\), and the angle we need to find (let's call it \(x\)) is an exterior angle at \(P\) with respect to the triangle. Wait, actually, we can also use the fact that the sum of angles in a triangle is \(180^{\circ}\) to find the interior angle at \(P\) first, then use linear pair. Let's do it that way.
First, find the interior angle at \(P\) (let's call it \(y\)) in triangle \(RQP\). We know that the sum of angles in a triangle is \(180^{\circ}\), so \(y + 35^{\circ}+95^{\circ}=180^{\circ}\).
Solving for \(y\): \(y=180^{\circ}-(35^{\circ} + 95^{\circ})=180^{\circ}-130^{\circ}=50^{\circ}\).
Then, the angle we need to find (let's call it \(x\)) and \(y\) form a linear pair, so \(x + y=180^{\circ}\). Since \(y = 50^{\circ}\), then \(x=180^{\circ}-50^{\circ}=130^{\circ}\). Wait, no, wait. Wait, the exterior angle theorem: the exterior angle at \(P\) (the angle we need to find) should be equal to the sum of the two non - adjacent interior angles. The two non - adjacent interior angles to the exterior angle at \(P\) are the angle at \(R\) (\(35^{\circ}\)) and the angle at \(Q\) (\(95^{\circ}\)). So by exterior angle theorem, the measure of the angle (let's call it \(x\)) is \(35^{\circ}+95^{\circ}=130^{\circ}\)? Wait, no, that can't be. Wait, no, the angle at \(Q\) is \(95^{\circ}\), which is an obtuse angle. Wait, maybe I misidentified the triangle. Wait, the angle at \(Q\) is \(95^{\circ}\), angle at \(R\) is \(35^{\circ}\), so the interior angle at \(P\) is \(180 - 35-95=50^{\circ}\), and then the angle we need to find is supplementary to this interior angle? Wait, no, the line \(PG\) and \(PQ\) form a linear pair? Wait, no, \(PG\) is a straight line extension of \(RP\)? Wait, no, looking at the diagram, \(R - P - G\) is a straight line? Wait, no, \(P\) is on \(RG\), and \(PQ\) is a segment from \(P\) to \(Q\), with \(RQ\) and \(PQ\) forming a right - angled - like? Wait, no, let's re - examine.

Wait, the correct approach: The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. The angle we need to find is an exterior angle for triangle \(RQP\), with the two non - adjacent interior angles being \(\angle R = 35^{\circ}\) and \(\angle RQP=95^{\circ}\). So the measure of the exterior angle (the angle we need to find) is \(35^{\circ}+95^{\circ}=130^{\circ}\)? Wait, no, that would mean the exterior angle is greater than \(180^{\circ}\)? No, that's wrong. Wait, no, the angle at \(Q\) is \(95^{\circ}\), which is an interior angle, and the angle at \(R\) is \(35^{\circ}\), so the exterior angle at \(P\) should be equal to the sum of the two non - adjacent interior angles. Wait, no, the sum of \(35 + 95=130\), and then the interior angle at \(P\) is \(50\), so the exterior angle (the angle we need to find) is \(180 - 50 = 130\)? Wait, yes, because the interior angle at \(P\) is \(50\), and the angle we need to find is supplementary to it (since they form a linear pair). So either way, using exterior angle theorem: exterior angle=\(35^{\circ}+95^{\circ}=130^{\circ}\), or using linear pair: \(180-(180 - 35 - 95)=35 + 95=130^{\circ}\).

Wait, no, I think I made a mistake. Wait, the angle at \(Q\) is \(95^{\circ}\), which is an interior angle, and the angle at \(R\) is \(35^{\circ}\), so the third interior angle (at \(P\)) is \(180-(35 + 95)=50^{\circ}\). Then the angle we need to find is adjacent to this \(50^{\circ}\) angle and forms a linear pair, so it is \(180 - 50=130^{\circ}\).

Answer:

\(130^{\circ}\)