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:

The answer is below

Step-by-step explanation:

In triangle DEF, ∠D = 30°, ∠F = 60° and EF = 6. Find DE + DF.

Solution:

A triangle is a polygon with three sides. There are different types of triangles such as isosceles triangle, scalene triangle equilateral triangle, right angled triangle.

In triangle DEF:

∠D + ∠E + ∠F = 180° (sum of angles in a triangle)

30 + ∠E + 60 = 180

∠E + 90 = 180

∠E = 180 - 90

∠E = 90°

This is a right angled triangle since one angle = 90° (∠E).

Using sine rule:

$\frac{|EF|}{sin(D)}=\frac{|DE|}{sin(F)} \\\\\frac{6}{sin(30)}=\frac{|DE|}{sin(60)} \\\\|DE|=10.39\\\\\frac{|EF|}{sin(D)}=\frac{|DF|}{sin(E)} \\\\\frac{6}{sin(30)}=\frac{|DF|}{sin(90)} \\\\|DF|=12$

Hence DE + DF = 12 + 10.39 = 22.39