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3 years ago

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Determine the approximate value of θ. A. 38.9 B. 51.1 C. 37.9 D. 52.1

1 answer:

Pepsi [2]3 years ago

8 0

Answer:

Option B. $51.1^o$

Step-by-step explanation:

we know that

In the right triangle of the figure

The sine of angle theta is equal to divide the opposite side to angle theta by the hypotenuse

so

$sin(\theta)=\frac{DF}{DE}$

substitute the given values

$sin(\theta)=\frac{7}{9}$

using a calculator

$\theta=sin^{-1}(\frac{7}{9})=51.1^o$

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Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

$\rho = \sqrt{x^{2}+y^{2}+z^{2}}$

$\phi = cos^{-1}\frac{z}{\rho}$

For angle θ:

If x > 0 and y > 0: $\theta = tan^{-1}\frac{y}{x}$ ;

If x < 0: $\theta = \pi + tan^{-1}\frac{y}{x}$ ;
If x > 0 and y < 0: $\theta = 2\pi + tan^{-1}\frac{y}{x}$ ;

Calculating:

a) (4,2,-4)

$\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}$ = 6

$\phi = cos^{-1}(\frac{-4}{6})$

$\phi = cos^{-1}(\frac{-2}{3})$

For θ, choose 1st option:

$\theta = tan^{-1}(\frac{2}{4})$

$\theta = tan^{-1}(\frac{1}{2})$

b) (0,8,15)

$\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}$ = 17

$\phi = cos^{-1}(\frac{15}{17})$

$\theta = tan^{-1}\frac{y}{x}$

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = $\frac{\pi}{2}$

c) (√2,1,1)

$\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}$ = 2

$\phi = cos^{-1}(\frac{1}{2})$

$\phi$ = $\frac{\pi}{3}$

$\theta = tan^{-1}\frac{1}{\sqrt{2} }$

d) (−2√3,−2,3)

$\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}$ = 5

$\phi = cos^{-1}(\frac{3}{5})$

Since x < 0, use 2nd option:

$\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }$

$\theta = \pi + \frac{\pi}{6}$

$\theta = \frac{7\pi}{6}$

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

$r=\sqrt{x^{2}+y^{2}}$

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

$r=\sqrt{4^{2}+2^{2}}$ = $\sqrt{20}$

$\theta = tan^{-1}\frac{1}{2}$

z = -4

b) (0, 8, 15)

$r=\sqrt{0^{2}+8^{2}}$ = 8

$\theta = \frac{\pi}{2}$

z = 15

c) (√2,1,1)

$r=\sqrt{(\sqrt{2} )^{2}+1^{2}}$ = $\sqrt{3}$

$\theta = \frac{\pi}{3}$

z = 1

d) (−2√3,−2,3)

$r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}$ = 4

$\theta = \frac{7\pi}{6}$

z = 3

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3 years ago