In the figure, $$ABC$$ and $$PQR$$ are two triangles such that $$\angle A \colon \angle B \colon \angle C = 5 \colon 6 \colon 7$$ and $$\angle PRQ = \angle B$$. $$PS$$ makes an angle $$\frac{\angle P}{3}$$ with $$PQ$$ and $$RS$$ makes an angle $$\frac{\angle SRT}{5}$$ with $$RQ$$. Then the measure of $$\angle S$$ is

We are given that $$∠A:∠B:∠C=5:6:7$$

So let us assume that $$∠A=5x$$°, $$∠B=6x$$° and $$∠C=7x$$°

Now, in triangle $$ABC$$

$$\angle A+\angle B+\angle C=180$$°

$$5x+6x+7x=180$$°

$$18x=180$$°

$$x=10$$°

So, $$\angle A=50$$°, $$\angle B=60$$° and $$\angle C=70$$°$$

Now, $$\angle PRQ=\angle B=60$$°

Now, since $$PRT$$ is a straight line and we were also given that $$\angle QRS=\dfrac{\angle{SRT}}{5}$$

$$\angle PRQ+\angle QRS+\angle SRT=180$$°

$$60+\dfrac{\angle{SRT}}{5}+\angle SRT=180$$°

From here, we get:

$$\angle SRT=100$$° and $$\angle SRQ=20$$° 

Now, in right triangle $$PQR$$, we have:

$$\angle RPQ+\angle PQR+\angle PRQ=180$$° 

$$\angle RPQ+90+60=180$$°

$$\angle RPQ=30$$°

Now, we were given that, $$\angle SPQ=\dfrac{RPQ}{3}$$

$$\angle SPQ=\dfrac{30}{3}=10$$°

And, $$\angle RPS=30-\angle SPQ=30-10=20$$°

Now, in triangle PSR, we have:

$$\angle RPS+\angle PSR+\angle SRP=180$$°

$$20+\angle PSR+(20+60)=180$$°

$$\angle PSR=180-100=80$$°

So, $$\angle S=80$$°