This is a special right triangle (30-60-90) and I can do most of them but this one confuses me how do you set up the problem?

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lazirlufias lazirlufias · Answer 1 · 2020-02-11T06:03:06+01:00

Answer:

[tex]x =\frac{40}{3}\sqrt{3} ; y =\frac{20}{3}\sqrt{3} [/tex]

Step-by-step explanation:

standart special right triangle with some square root operations

let the triangle has side a, 2a, and a√3.

since a√3 = 20, then [tex]a=\frac{20}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{20}{3}\sqrt{3}[/tex]

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Favouredlyf Favouredlyf · Answer 2 · 2020-02-11T07:10:21+01:00

Answer:

Step-by-step explanation:

With 60 degrees as the reference angle,

x represents the hypotenuse of the right angle triangle.

y represents the adjacent side of the right angle triangle

20 represents the opposite side of the right angle triangle.

To determine x, we would apply the

Sine trigonometric ratio. It is expressed as

Sin θ = opposite side/hypotenuse.

Sin 60 = 20/x

x = 20/Sin60 = 20/0.8660

x = 23.1

To determine y, we would apply the

Tangent trigonometric ratio. It is expressed as

Tan θ = opposite side/adjacent side.

Tan 60 = 20/y

y = 20/Tan60 = 20/1.7321

y = 11.5