Question

In triangle $DEF,$ $\angle D = 30^\circ,$ $\angle F = 60^\circ,$ and $EF = 6.$ Find $DE + DF.$ [asy] unitsize(2 cm); pair D, E, F; D = (0,sqrt(3)); E = (0,0); F = (1,0); draw(D--E--F--cycle); label("$D$", D, N); label("$E$", E, SW); label("$F$", F, SE); label("$6$", (E + F)/2, S); [/asy]

Answer 1

Answer:

Step-by-step explanation:

The answer to this question is 12+6sqrt3.

This is because of the 30, 60, and 90 degree rule. The rule states that the hypotenuse is equal to twice the length of the shorter leg, which is the side across from the 30 degree angle.

Using the rule you now know that DF is 12. Then use the Pythagorean Theorem to find the length of DE. 12^2=144, and 6^2=36. 144-36=108.

The square root of 108 simplified=6sqrt3.

Then you add them together and get 12+6sqrt3.