Solve for X

Mathematics

1 answer:

Phantasy [73]3 years ago

4 0

Answer:

x = 48

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

156° is an exterior angle of the triangle, then

108 + x = 156 ( subtract 108 from both sides )

x = 48

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Which expression is equivalent to (sqrt3 + i)^3?

Anarel [89]

Express √3 + i in polar form:

|√3 + i| = √((√3)² + 1²) = √4 = 2

arg(√3 + i) = arctan(1/√3) = π/6

Then

√3 + i = 2 (cos(π/6) + i sin(π/6))

By DeMoivre's theorem,

(√3 + i)³ = (2 (cos(π/6) + i sin(π/6)))³

… = 2³ (cos(3 • π/6) + i sin(3 • π/6))

… = 8 (cos(π/2) + i sin(π/2))

… = 8i

4 0

3 years ago

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$