Lines $P S, Q T$ and $R U$ intersect at a common point $O$, as shown. $P$ is joined to $Q, R$ to $S$, and $T$ to $U$, to form triangles. The value of $\angle P+\angle Q+\angle R+\angle S+\angle T+\angle U$ is
(A) $450^{\circ}$
(D) $540^{\circ}$
(B) $270^{\circ}$
(E) $720^{\circ}$
(C) $360^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-093.jpg?height=299&width=480&top_left_y=707&top_left_x=1321)
Lines $P S, Q T$ and $R U$ intersect at a common point $O$, as shown. $P$ is joined to $Q, R$ to $S$, and $T$ to $U$, to form triangles. The value of $\angle P+\angle Q+\angle R+\angle S+\angle T+\angle U$ is
(A) $450^{\circ}$
(D) $540^{\circ}$
(B) $270^{\circ}$
(E) $720^{\circ}$
(C) $360^{\circ}$

Solution

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-414.jpg?height=309&width=501&top_left_y=1144&top_left_x=1343)

We know that $\angle P O Q+\angle P O U+\angle U O T=180^{\circ}$ because when added they form a straight line. Since $\angle P O U=\angle R O S$ (vertically opposite angles), therefore $\angle P O Q+\angle R O S+\angle U O T=180^{\circ}$.

Thus the sum of the remaining angles is just $3 \times 180^{\circ}-180^{\circ}=360^{\circ}$ since there are $180^{\circ}$ in each of the three given triangles.

ANswer: (C)