Answer:
[tex]ST=11.2\:\mathrm{m}[/tex]
Step-by-step explanation:
Since [tex]\angle T[/tex] corresponds with [tex]\angle Q[/tex], we need to find the measure of [tex]\angle Q[/tex] to set up a proportion.
In [tex]\triangle PQR[/tex], two sides and the angle between them are given, so we can use the Law of Cosines to find the third side.
The Law of Cosines is given as:
[tex]c^2=a^2+b^2-2ab\cos C[/tex]
Plugging in values, we get:
[tex]c^2=8^2+7^2-2\cdot8\cdot 7\cdot \cos 59^{\circ},\\c\approx 7.44[/tex].
Now we can use the Law of Sines to find [tex]m\angle Q[/tex]:
[tex]\frac{\sin59^{\circ}}{7.44}=\frac{\sin Q}{7},\\Q=\arcsin( \frac{7\cdot \sin 59^{\circ}}{7.44})\approx 53.75^{\circ}[/tex].
Using this, we can set up the following proportion to solve for [tex]ST[/tex]:
[tex]\frac{53.75}{75}=\frac{8}{ST},\\53.75\cdot ST=8\cdot 75,\\ST=\frac{8\cdot 75}{53.75}\approx \fbox{$11.2\:\mathrm{m}$}[/tex].
