MATH SOLVE

3 months ago

Q:
# the height h of the equilateral triangle below is given by y= 5 cot theta where theta = 30 degreesA) 2.9B)4.3C)7.1D)8.7

Accepted Solution

A:

To solve this we can take two different approaches:

1. We can substitute theta by 30° and evaluate our expression using a calculator:

[tex]y=5cot( \alpha )[/tex]

[tex]y=5cot(30)[/tex]

[tex]y=8.7[/tex]

2. We can use the unitary circle and the fact that [tex]cot \alpha = \frac{cos( \alpha }{sin( \alpha )} [/tex], so we can rewrite our expression as follows:

[tex]y=5cot( \alpha )[/tex]

[tex]y=5 \frac{cos( \alpha )}{sin( \alpha )} [/tex]

[tex]y= \frac{cos(30)}{sin(30)} [/tex]

From our unitary circle we can check that [tex]cos(30)= \frac{ \sqrt{3} }{2} [/tex] and [tex]sin(30)= \frac{1}{2} [/tex].

Lets replace those values in our expression and simplify:

[tex]y= \frac{ 5(\frac{ \sqrt{3} }{2})}{ \frac{1}{2} }[/tex]

[tex]y=5 \sqrt{3} [/tex]

[tex]y=8.7[/tex]

Either way we can conclude that the correct answer is: D)8.7

1. We can substitute theta by 30° and evaluate our expression using a calculator:

[tex]y=5cot( \alpha )[/tex]

[tex]y=5cot(30)[/tex]

[tex]y=8.7[/tex]

2. We can use the unitary circle and the fact that [tex]cot \alpha = \frac{cos( \alpha }{sin( \alpha )} [/tex], so we can rewrite our expression as follows:

[tex]y=5cot( \alpha )[/tex]

[tex]y=5 \frac{cos( \alpha )}{sin( \alpha )} [/tex]

[tex]y= \frac{cos(30)}{sin(30)} [/tex]

From our unitary circle we can check that [tex]cos(30)= \frac{ \sqrt{3} }{2} [/tex] and [tex]sin(30)= \frac{1}{2} [/tex].

Lets replace those values in our expression and simplify:

[tex]y= \frac{ 5(\frac{ \sqrt{3} }{2})}{ \frac{1}{2} }[/tex]

[tex]y=5 \sqrt{3} [/tex]

[tex]y=8.7[/tex]

Either way we can conclude that the correct answer is: D)8.7