First we are given that △ EQU is equilateral, so it has three angles that have a measure of 60^(∘). Notice that the height in a right triangle divides it into two 30^(∘)-60^(∘)-90^(∘) triangles.
If we call the length of ES a, then the length of EQ is 2a and the length of QS is asqrt(3).
Using these lengths we can evaluate the sine of ∠ E, which is 60^(∘). Recall that the sine of an angle is the ratio between the length of the leg opposite this angle to the length of the hypotenuse.
The value of the sine of ∠ E is sqrt(3)2. Now let's move to the second triangle. We are given that △ RGT is a right triangle with RG=2, RT= 1, and ∠ T is a right angle. Let's draw this triangle.
Notice that when the length of the hypotenuse in a right triangle is two times the length of the shorter leg this triangle is a 30^(∘)-60^(∘)-90^(∘) triangle. Therefore m∠ R is 60^(∘) and m∠ G is 30^(∘). Additionally, the length of GT is sqrt(3).
Now we can evaluate the cosine of ∠ G, which has a measure of 30^(∘). Let's recall that in a right triangle the cosine of an angle is the ratio between the length of the leg adjacent to this angle and the hypotenuse.
cos 30^(∘)=sqrt(3)/2
The value of the cosine of ∠ G is also sqrt(3)2. Therefore the sine of ∠ E is equal to the cosine of ∠ G. Notice that in case of supplementary angles the sine of one of them is always equal to the cosine of the other one.
