In a 30 ^(∘) - 60 ^(∘) - 90 ^(∘) triangle, the hypotenuse h is 2 times the length of the shorter leg s and the longer leg l is sqrt(3) times the length of the shorter leg.
Let's name the vertices of the given figure. Then we can find y and x, one at a time.
Finding y
We are given that ∠BEC is a right angle and that EC and EB are congruent. Therefore, △ BCE is a 45^(∘) - 45 ^(∘) - 90 ^(∘) triangle. We also know that the length of the leg EC is 5 sqrt(2). By the 45^(∘) - 45 ^(∘) - 90 ^(∘) Triangle Theorem, l= 5 sqrt(2) and the hypotenuse y is sqrt(2) times the length of a leg, l.
y=l sqrt(2)
To find y, let's substitute l= 5 sqrt(2) in this equation.
We are given that ∠BCD is a right angle, and through the process of finding y, we know that the measure of ∠BCE is 45 ^(∘). Since ∠DCE and ∠BCE are complementary, the measure of ∠DCE is also 45^(∘). Therefore, △ ECD is also a 45^(∘) - 45 ^(∘) - 90 ^(∘) triangle.
The length of the leg is x and the length of the hypotenuse is 5 sqrt(2). By the 45^(∘) - 45 ^(∘) - 90 ^(∘) Triangle Theorem, the length of the hypotenuse is sqrt(2) times the length of a leg, x.
x sqrt(2) = 5 sqrt(2) ⇔ x=5
