Math Made Easy: Problem of the Day 101

Diagram, diagram, diagram. Always draw a picture when you can. It will clear things up and make sure you get it right. Given that all the vertices of our triangles correspond, we’re able to make our diagram look like this:

From that, the rest is easy. We see that Angle P is also a right angle. DQ is the hypotenuse of Triangle PDQ. We remind ourselves that the sine of a right triangle is just the length of the side opposite that angle over the length of the hypotenuse. So we first need to find the length of DQ. The Pythagorean Theorem helps us out there – it’s just the square root of the sum of the squares of the other two sides.

Once we have that, we just plug the lengths into our expression for the sine of Angle D, and we’re actually done.

We would simplify if we could, but that’s already in simplest terms.

Today’s problem was very fast, and relatively simple. Its difficulty lies mostly in a word soup that sometimes distracts people from being able to solve word problems. But when we diagram, we cut through the word soup and help to clarify exactly what the situation being described is. So do it! Too many people resist drawing and try to juggle everything in their heads when, if they’d just take the few seconds (okay, maybe a minute or two) to do their diagram, things would become much easier.

To help keep my site free, please become a patron at my Patreon:

Or you can make a one-time contribution at my Paypal: