IS my answer right ?????????????
Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. B = 46°, a = 12, b = 11
A = 51.7°, C = 82.3°, c = 15.2; A = 128.3°, C = 5.7°, c = 1.5
@Anu2401 @e.mccormick @cwrw238 @Falco276 @IrishBoy @ivandelgado @johnweldon1993 @kropot72 @ogkat @Preetha @RaphaelFilgueiras @thomaster
can any help
First OK. Still checking second.
ok thnx
Helps when I put the riht numbers in the calculator.... looks good.
wow thnx i have one more that im confuse about @e.mccormick if you dont mind
Find the fourth roots of 256(cos 280° + i sin 280°).
DeMoivre’s Theorem Let \(z = r[\cos(θ) + i\sin(θ)]\) and n be a positive integer. Then \(z^n=r^n[\cos(n\theta)+i\sin(n\theta)]\)
Let y=256(cos 280+i sin 280) Now this[(cos 280+i sin 280)] can be written in exponential form as exp^(i*280) We need to find y^(1/4) Hence =>[256*exp^(280i)]^(1/4) =>4*exp^(70i) =>4(cos70+isin70) Ans
i dont understand the formula the second one what is r 256
So you do your \(256^4\) and put that out front, then multiply the angles by 4, then simplify a bit (can reduce the angles to \(0^\circ\le 360^\circ\)
Oops. Yah, root.... I want the wrong direction.
ok @Anu2401 what is after that
Anu2401's /(r^{\frac{1}{4}} is right. Going to the root rather than the power.
\(r^{\frac{1}{4}}\)
so is he right at that step
im trying to take notes
@Easycheddar Yes the solution is correct . Just go through de-moiver's theorem once .... Google It :)
Absolutly. Because they said 4th root. It is the same thing as this:\[\sqrt[4]{x}=x^{\frac{1}{4}}\]
ok so that the solution now i just need to do the therom
I went tot he 4th power in my explanation... the question aid 4th root. My mistake there.
so what the answer cause i am getting something crazy @e.mccormick and @Anu2401
U can put the value of cos 70 & Sin 70 in last step . Use a calculator to find the values of those
Did you start with: \[256^{\frac{1}{4}}\left(\cos \frac{280^\circ}{4} + i \sin \frac{280^\circ}{4}\right)\]And follow how Anu2401 did it? Also, is your calculator in Degrees or Radians mode?
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