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ANEK [815]
3 years ago
5

I need trigometry help

Mathematics
1 answer:
Alex Ar [27]3 years ago
4 0
A^2=b^2+c^2-2bc Cos(A)
(270)^2=(255)^2+(442.85)^2-2(255)(442.85)Cos(A)
A=Cos^(-1)(270^2-255^2_(442.85)^2)/(-2*255*442.85)=33.5435
solve for angle A approximately is 33.54 degrees.
sinA/a=SinB/b
Sin33.54/270=SinB/255
B=Sin^(-1)(sin33.54/270*255)approximately is 31.45.
180-31.45-33.54=115.01 is Angle C.

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Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2
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Answer:

Proof below.

Step-by-step explanation:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

<u>Given quadratic equation</u>:

5x^2-6x-2=0

<u>Define the variables</u>:

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<u>Substitute</u> the defined variables into the quadratic formula and <u>solve for x</u>:

\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}

\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}

\implies x=\dfrac{6 \pm \sqrt{76}}{10}

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\implies x=\dfrac{3 \pm \sqrt{19}}{5}

Therefore, the exact solutions to the given <u>quadratic equation</u> are:

x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}

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