Finding the secant from an angle that isn't cosine

[wduk]

Hello

I was hoping some one could check if i have understood this correctly.

I have an angle of sin(theta) = 10/45

From this i can calculate that the adjacent must be of length 5sqrt(77). So to find the secant i can do:

sec(theta) = 1 / (5sqrt(77)/45)

Thus sec(theta) = 45/5sqrt(77)

My question though is, if i am given the angle sine(theta) = 10/45 should i just be assuming it is a right angled triangle like i did in order to find the length of the adjacent? Have i understood this correctly? I keep doubting myself as having understood it correctly.

 

[Deleted member 4993]

Hello

I was hoping some one could check if i have understood this correctly.

I have an angle of sin(theta) = 10/45

From this i can calculate that the adjacent must be of length 5sqrt(77). So to find the secant i can do:

sec(theta) = 1 / (5sqrt(77)/45)

Thus sec(theta) = 45/5sqrt(77)

My question though is, if i am given the angle sine(theta) = 10/45 should i just be assuming it is a right angled triangle like i did in order to find the length of the adjacent? Have i understood this correctly? I keep doubting myself as having understood it correctly.

Your logic is correct if 0 ≤ Θ ≤ π/2

Another way to look at it:

sin²(Θ) = (10/45)² → 1 - cos²(Θ) = (10/45)² → cos²(Θ) = 1 - (10/45)²

cos(Θ) = ± √(1925/2025)

sec(Θ) = ± √(2025/1925)

 

[wduk]

How are you getting those numbers?

1 - (10/45)^2 = 77/81 is what i get.

 

[ksdhart2]

Yes, the value you give is correct. However, the numbers Subhotosh Khan gives are also correct. His numbers are just yours multiplied by 25: 1925/2025 = (77 * 25)/(81 * 25) = 77/81.

 

[wduk]

Is there a reason for multiplying by 25 though ?

 

[ksdhart2]

I don't think the specific number 25 has any significance, just a coincidence. Most likely, what happened was that he squared the term first, then subtracted. 1 - (10/45)² = 1 - (100/2025) = 2025/2025 - 100/2025 = 1925/2025

 

[pka]

My question though is, if i am given the angle sine(theta) = 10/45

Recall that \(\displaystyle \sin(\theta)=\frac{y}{r}\) thus \(\displaystyle y=10,~r=45~\&~x=\sqrt{(45)^2-(10)^2}=5\sqrt{77}\)
\(\displaystyle \begin{align*}\cos(\theta)&=\frac{x}{r} \\\tan(\theta)&=\frac{y}{x}\\\sec(\theta)&=\frac{r}{x} \\\csc(\theta)&=\frac{r}{y}\end{align*}\)

You must adjust for quadrant.