# Problem of the Day: 1/25/13

A photograph has a length 2 inches greater than its width. Its frame has a 2-inch border and a rectangular area of 120 square inches. What is the width of the photograph?

Solution to yesterday’s problem:

By supplementary angles, we know that angle DBC is 40°. Then, because angles in a triangle add up to 180°, we can say that $D + B=140$. We are also given that angle B is 80 degrees more than angle D, so we can represent angle B as $D+80$. Then we have $D+D+80=140$. Solving this equation,

$2D=60$

$D=30$

We conclude that angle D = 30°, angle B = 80°, and angle DBC = 40°. This checks out because the sum of the angles is 180°.