In △DCE, DC = CE = 6, m∠E = 53º. Find DE.
So I tried to use tan or cot but there isn't a 90 degree angle C is 74 i somehow got 7.22 but i can't really explain how.
since the triangle is isosceles,
m∠D = m∠E = 53º
so, m∠C = 74º
so, (DE/2)/6 = sin(74/2)º
DE/12 = sin37º
DE = 12 sin37º = 7.22
or, using the law of sines,
DE/sin74º = 6/sin53º
DE = 7.22
or, using the law of cosines,
DE^2 = 6^2+6^2 - 2*6*6 cos74º
DE = 7.22
As you can see, the key to this was getting angle C
Thank you!
Thanks so much
To find DE in triangle DCE, we can use the Law of Cosines, since we know two sides and an angle.
The Law of Cosines states:
c^2 = a^2 + b^2 - 2ab * cos(C)
In triangle DCE, we have:
- DC = 6
- CE = 6
- Angle E = 53º
- We want to find DE
Using the Law of Cosines, we have:
DE^2 = DC^2 + CE^2 - 2 * DC * CE * cos(E)
Substituting the given values, we get:
DE^2 = 6^2 + 6^2 - 2 * 6 * 6 * cos(53º)
Simplifying further:
DE^2 = 36 + 36 - 72 * cos(53º)
To calculate cos(53º), we can use a scientific calculator:
cos(53º) ≈ 0.5981
Substituting this value, we have:
DE^2 = 36 + 36 - 72 * 0.5981
Simplifying:
DE^2 ≈ 72 - 42.8707
DE^2 ≈ 29.1293
To find DE, we take the square root of both sides:
DE ≈ √29.1293
Using a calculator, we find:
DE ≈ 5.3947
Therefore, the approximate value of DE is 5.3947.