Let's add this information to our diagram. Let y represent the length of BC and z represent the length of DC.
Since △ ACD and △ BCD are right triangles, we can use one of the trigonometric ratios to solve for z and y. Let's recall that the tangent of ∠A is the ratio of the leg opposite ∠A to the leg adjacent ∠A. Using this definition, we can create the equations for tan 20^(∘) and tan 15^(∘).
tan 20^(∘)=z/y & (I) tan 15^(∘)=z/2+y & (II)
As we can see, we need to solve the system of equations. To do this, we can use the Substitution Method. Our first step will be to isolate z in the first equation.
The value of z is approximately 2. Let's add information we found to our diagram.
Finally we can find the value of x. To do this let's recall that the sine of an acute angle in a right triangle is a ratio of the leg opposite this angle to the hypotenuse of this triangle. Using this information, we will create an equation.
sin 15^(∘)=2/x
Let's solve the above equation.
