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# Help and much appreciated

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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

Jul 20, 2019

#1
+8829
+5

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.

Jul 20, 2019

#1
+8829
+5

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.

hectictar Jul 20, 2019