A portion of a power line support tower is to be constructed as congruent triangles, as shown on the diagram. The crossing beams will be 11m long and will intersect at an acute angle of 50 degrees, and angle S= 37 degrees. Find the lengths of all sides of the triangle QPT
https://gyazo.com/320d833e95efb64d446cd9b6277441ff -diagram
+9466 
By the SAS congruence theorem, △PQT ≅ △SRT So...
m∠QPT = m∠RST = 37°
m∠QTP = 180° - 50° = 130°
m∠PQT = 180° - 37° - 130° = 13°
Since QT = RT we can substitute QT in for RT in the next equation.
\(RT+TP\ =\ 11\\~\\ QT+TP\ =\ 11\\~\\ QT\ =\ 11 - TP\)
Now we can substitute 11 - TP in for QT in the next equation.
By the Law of Sines,
\( \frac{TP}{\sin13^\circ} \ =\ \frac{QT}{\sin37^\circ}\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ QT\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ 11-TP\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}+TP\ =\ 11\\~\\ TP\cdot(\frac{\sin37^\circ}{\sin13^\circ}+1)\ =\ 11\\~\\ TP\ =\ 11\div(\frac{\sin37^\circ}{\sin13^\circ}+1)\\~\\ TP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ} \)
Now that we know the length of TP , we can find the length of QT.
\(QT\ =\ 11-TP\\~\\ QT\ =\ 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\)
Finally, we can use the Law of Sines again to find QP.
\(\frac{QP}{\sin130^\circ}\ =\ \frac{TP}{\sin13^\circ}\\~\\ QP\ =\ TP\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}\)
So we have found:
\(\begin{array}{c} TP&=& \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 2.993\\~\\ QT& =& 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 8.007\\~\\ QP& =&\frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 10.192 \end{array}\)
And all lengths are in meters.
+9466
By the SAS congruence theorem, △PQT ≅ △SRT So...
m∠QPT = m∠RST = 37°
m∠QTP = 180° - 50° = 130°
m∠PQT = 180° - 37° - 130° = 13°
Since QT = RT we can substitute QT in for RT in the next equation.
\(RT+TP\ =\ 11\\~\\ QT+TP\ =\ 11\\~\\ QT\ =\ 11 - TP\)
Now we can substitute 11 - TP in for QT in the next equation.
By the Law of Sines,
\( \frac{TP}{\sin13^\circ} \ =\ \frac{QT}{\sin37^\circ}\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ QT\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ 11-TP\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}+TP\ =\ 11\\~\\ TP\cdot(\frac{\sin37^\circ}{\sin13^\circ}+1)\ =\ 11\\~\\ TP\ =\ 11\div(\frac{\sin37^\circ}{\sin13^\circ}+1)\\~\\ TP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ} \)
Now that we know the length of TP , we can find the length of QT.
\(QT\ =\ 11-TP\\~\\ QT\ =\ 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\)
Finally, we can use the Law of Sines again to find QP.
\(\frac{QP}{\sin130^\circ}\ =\ \frac{TP}{\sin13^\circ}\\~\\ QP\ =\ TP\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}\)
So we have found:
\(\begin{array}{c} TP&=& \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 2.993\\~\\ QT& =& 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 8.007\\~\\ QP& =&\frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 10.192 \end{array}\)
And all lengths are in meters.
