Answer: \(\mathbf p=-15\mathbf i+36\mathbf j\), \(\mathbf q=30\mathbf i-16\mathbf j\); \(|\mathbf p+\mathbf q|=25\), angle \(53.1^\circ\).

The direction vector for \(\mathbf p\) is

\(-5\mathbf i+12\mathbf j.\)

Its magnitude is

\(\sqrt{(-5)^2+12^2}=\sqrt{169}=13.\)

Since \(\mathbf p\) has magnitude \(39\), the scale factor is

\(\frac{39}{13}=3.\)

Therefore

The direction vector for \(\mathbf q\) is

\(15\mathbf i-8\mathbf j.\)

Its magnitude is

\(\sqrt{15^2+(-8)^2}=\sqrt{289}=17.\)

Since \(\mathbf q\) has magnitude \(34\), the scale factor is

\(\frac{34}{17}=2.\)

Therefore

Now

Its magnitude is

\(\sqrt{15^2+20^2}=\sqrt{625}=25.\)

If \(\alpha\) is the angle made with the positive \(x\)-axis, then

\(\tan\alpha=\frac{20}{15}=\frac43.\)

Hence

\(\alpha=\tan^{-1}\frac43=53.1^\circ\)

to 1 decimal place.