In triangle DEF measurement angle D = 45 degrees meausrement angle E = 63 degrees and EF = 24 inches. What is DE to the nearest tenth of a inch?
since the angles add to 180°, angle F is 72°.
Using the law of sines,
DF/sin72° = 24/sin45°
x = 32.28
So, did you just guess at A?
This is just an answer from another website, but it should still work.
Answer:
DE = 32.8 inches is the answer.
Step-by-step explanation:
It is given that In the given triangle DEF, angle D = 45°, ∠E = 63° and EF = 24 inches.
We have to calculate the length of DE.
From the sine rule of a triangle
\(\frac{sin D}{EF}=\frac{sinF}{DE}\)
As we know sum of all angles of a triangle is 180°
Therefore ∠D + ∠E + ∠F = 180°
45 + 63 + F = 180
F = 180-108 = 72°
Now from the sine rule written above
\(\frac{sin45}{24}=\frac{sin72}{DE}\)
\(\frac{0.71}{24}=\frac{0.95}{DE}\)
.029 = .95/DE
DE = 0.95/0.029 = 32.75 ≈ 32.80 inches
Answer will be DE = 32.80 inches