$J K L M$ is a square. Points $P$ and $Q$ are outside the square such that triangles $J M P$ and $M L Q$ are both equilateral. The size, in degrees, of angle $P Q M$ is
(A) 10
(D) 30
(B) 15
(C) 25
(E) 150

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-102.jpg?height=315&width=352&top_left_y=794&top_left_x=1342)
$J K L M$ is a square. Points $P$ and $Q$ are outside the square such that triangles $J M P$ and $M L Q$ are both equilateral. The size, in degrees, of angle $P Q M$ is
(A) 10
(B) 15
(C) 25
(D) 30
(E) 150

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-445.jpg?height=458&width=439&top_left_y=232&top_left_x=1388)

Solution

$\angle P M Q=360^{\circ}-90^{\circ}-60^{\circ}-60^{\circ}=150^{\circ}$

Since $\triangle P Q M$ is isosceles, $\angle P Q M=\frac{180^{\circ}-150^{\circ}}{2}=15^{\circ}$.

ANSWER: (B)