In cyclic quadrilateral \(PQRS\)
\(\frac{\angle P}{3} = \frac{\angle Q}{4} = \frac{\angle R}{5}\)
Find the largest angle in quadrilateral in degrees.
+118608 Opposite angles in a cyclic quad are suplementar
Givew those equalities I can see that
The smallest would add with the biggest to get 180 degrees
The largest angle could be
\(\frac{\angle q}{4}=\frac{\angle r}{5}\\ \angle q=0.8*\angle r\\~\\ \angle q + \angle r =180^ \circ\\ 0.8*\angle r + \angle r =180^ \circ\\ 1.8*\angle r =180^ \circ\\ \angle r =100^ \circ\\ \angle q = 80 ^\circ\\~\\ \frac{\angle p}{3}=\frac{\angle r}{5}\\ \frac{\angle p}{3}=\frac{100}{5}\\ \angle p = 60^\circ\\ so\\ \angle s = 120 ^\circ \)
LaTex:
\frac{\angle q}{4}=\frac{\angle r}{5}\\
\angle q=0.8*\angle r\\~\\
\angle q + \angle r =180^ \circ\\
0.8*\angle r + \angle r =180^ \circ\\
1.8*\angle r =180^ \circ\\
\angle r =100^ \circ\\
\angle q = 80 ^\circ\\~\\
\frac{\angle p}{3}=\frac{\angle r}{5}\\
\frac{\angle p}{3}=\frac{100}{5}\\
\angle p = 60^\circ\\
so\\
\angle s = 120 ^\circ
