Geometry and Trigonometry and me.

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Denzel

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I am confused. Since sinθ = √3 ÷ 2 , using the Pythagorean theorem to find for the missing side, I get the length as 1÷2

I know that the answer of (a) (i) = 1÷2 I don't know what steps to take to get it though.

And since we have all sides of the triangle then, tanθ = 2 ÷ 1/2, isn't it? but for (a) (ii) how is the Ans= √3


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Given that \(\displaystyle \sin(\theta)=\frac{\sqrt3}{2}\)
\(\displaystyle \begin{align*}1&=\sin^2+\cos^2(\theta) \\&=\frac{3}{4}+\cos^2(\theta)\\\frac{1}{4}&=\cos^2(\theta)\\\frac{1}{2}&=\cos(\theta) \end{align*}\)

Can you finish?
 
Denzel said:
I am confused. Since sinθ = √3 ÷ 2 , using the Pythagorean theorem to find for the missing side, I get the length as 1÷2

I know that the answer of (a) (i) = 1÷2 I don't know what steps to take to get it though.

And since we have all sides of the triangle then, tanθ = 2 ÷ 1/2, isn't it? but for (a) (ii) how is the Ans= √3


View attachment 11984
Click to expand...
I assume that you are working only with acute angles; otherwise this problem would not have unique answers. So your working with a right triangle is appropriate.

But you haven't named the sides, and seem to have mixed up some details.

If sin θ = √3 / 2, we can construct a triangle whose opposite leg is √3, and whose hypotenuse is 2. By the Pythagorean Theorem, the adjacent leg is then 1.

Then the cosine of θ is the adjacent leg, 1, over the hypotenuse, 2, making it 1/2.

But the tangent is the opposite leg, √3, over the adjacent leg, 1, making it √3.

I think you did it a little differently, and had opposite leg = √3 / 2, hypotenuse = 1, and adjacent leg = 1/2. But then the tangent is (√3 / 2)/(1/2) = √3. Either way, you seem to have used the wrong sides for the tangent. No side, in your version, is 2.

I'm assuming you meant to talk about (a)(i)(a and b), not (a)(i) and (a)(ii).
 
pka said:
Given that \(\displaystyle \sin(\theta)=\frac{\sqrt3}{2}\)
\(\displaystyle \begin{align*}1&=\sin^2+\cos^2(\theta) \\&=\frac{3}{4}+\cos^2(\theta)\\\frac{1}{4}&=\cos^2(\theta)\\\frac{1}{2}&=\cos(\theta) \end{align*}/tex]
\)
Click to expand...

Denzel said:
I tried but I didn't get 1. How?
Click to expand...
\(\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\) AND \(\displaystyle \frac{\sin(\theta)}{\tan(\theta)}=~???\)

Why can you not finish?
 
pka said:


\(\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\) AND \(\displaystyle \frac{\sin(\theta)}{\tan(\theta)}=~???\)

Why can you not finish?
Click to expand...
Oh, sorry. I understood how you got 1/2. I was trying to figure out how Dr. Peterson got 1 as an answer, using the Pythagorean Theorem but can you explain how you got this formula?
 
I can finish though!

tan(θ) = sin(θ) / cos(θ)
tan(θ) = √3/2 ÷ 1/2
tan(θ) = √3/2 * 2
tan(θ) = √3
 
Oh. I see my mistake. Adj =1 / Hyp =2

Dr.Peterson said:
I assume that you are working only with acute angles; otherwise this problem would not have unique answers. So your working with a right triangle is appropriate.

But you haven't named the sides, and seem to have mixed up some details.

If sin θ = √3 / 2, we can construct a triangle whose opposite leg is √3, and whose hypotenuse is 2. By the Pythagorean Theorem, the adjacent leg is then 1.

Then the cosine of θ is the adjacent leg, 1, over the hypotenuse, 2, making it 1/2.

But the tangent is the opposite leg, √3, over the adjacent leg, 1, making it √3.

I think you did it a little differently, and had opposite leg = √3 / 2, hypotenuse = 1, and adjacent leg = 1/2. But then the tangent is (√3 / 2)/(1/2) = √3. Either way, you seem to have used the wrong sides for the tangent. No side, in your version, is 2.

I'm assuming you meant to talk about (a)(i)(a and b), not (a)(i) and (a)(ii).
Click to expand...
 
What you meant may be correct, but what you wrote is not.

You wrote correctly that [MATH]\sqrt{b^2} = \sqrt{1}[/MATH]. But when you evaluate each side, you don't get [MATH]b = \frac{1}{2}[/MATH]; you get [MATH]b = 1[/MATH]! Then the tangent is [MATH]\frac{a}{b} = \frac{\sqrt{3}}{1} = \sqrt{3}[/MATH].
 
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