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stepladder [879]

3 years ago

8

A solid oblique pyramid has a regular hexagonal base with an area of 54StartRoot 3 EndRoot cm2 and an edge length of 6 cm. Angle

BAC measures 60°. What is the volume of the pyramid?

2 answers:

Klio2033 [76]3 years ago

4 0

Answer:

IT's C

Step-by-step explanation:

I take the quiz

exis [7]3 years ago

3 0

Answer:

its c 324cm3

Step-by-step explanation:

i took the test

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Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y =

Black_prince [1.1K]

Answer:

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

Step-by-step explanation:

First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line $x + 2y= 0, x\geq 0$

If we choose $x = 2$ (we can do so because of the condition $x \geq 0$ , which means that any positive value is suitable for $x$ ), then we have

$2 +2y = 0\implies 2 = -2y \implies y = -1$

Therefore, the terminal side of the angle $\theta$ is passing through the origin and the point $(2,-1)$ and now we can draw it.

The angle $\theta$ is presented below.

The distance of the point $(2,-1)$ from the origin equals

$r = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

Now, we can determine the values of the six trigonometric function, by using their definitions.

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

6 0

3 years ago