Sun Local Hour Angle and Latitude

moja_a

1. Homework Statement

if:
H = Local Hour Angle
Lat = Latitude.
Dec = Sun Declination.

cos(H) = -sin(a)-sin(Lat)*sin(Dec) / cos(Lat)*cos(Dec)

I wand to get The value of Lat .

3. The Attempt at a Solution
I Tried to make it simple By :
1 - multiply both sides by the denominator
1- square both sides
2- Rewrite cos^2(L) as 1-sin^2(L)
3- Rearrange into a quadratic equation in sin(L).
And this is what I got:

Cos^2(H).Cos^2(D).Cos^2(L) = -Sin^2(a) - 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D)

Cos^2(H).Cos^2(D).(1-Sin^2(L) )= -Sin^2(a) - 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D)

Cos^2(H).Cos^2(D) - Cos^2(H).Cos^2(D).Sin^2(L) = - Sin^2(a) - 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D)

Cos^2(H).Cos^2(D) = -Sin^2(a)- 2Sin(a).Sin(D).Sin(L) + Sin^2(L).Sin^2(D) + Cos^2(H).Cos^2(D).Sin^2(L)

Cos^2(H).Cos^2(D) = -Sin^2(a)- 2Sin(a).Sin(D).Sin(L) + Sin^2(L) . ( Sin^2(D) + Cos^2(H).Cos^2(D) )

then I am Stuck can you help me please.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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Redbelly98

Staff Emeritus
Homework Helper
Welcome to Physics Forums.

So, is the problem asking you to solve the equation for Lat (or "L")?

As I read the equation, Lat appears in the form

sin(Lat) / cos(Lat)​

This can be replaced by a different, well-known trig function.

moja_a

Yes I used L = Lat
can you tell me about the function that replace it

Last edited:

moja_a

can you tell me about the function that replace it

Redbelly98

Staff Emeritus
Homework Helper
There are six standard, well-known trig functions. It is the one that is equivalent to

sin(θ) / cos(θ) = ___?​

Surely you know this?

moja_a



may be I wrote the formula incorrectly

$$Cos H = \frac{sin(a)-sin(L)*Sin(D)}{Cos(L)*Cos(D)}$$

I want to get the value of sin(L)

gneill

Mentor
$$cos(H) = \frac{sin(a)-sin(L)*sin(D)}{cos(L)*cos(D)}$$

$$cos(L) = \frac{sin(a)-sin(L)*sin(D)}{cos(H)*cos(D)}$$

$$cos(L) = \frac{sin(a)}{cos(H)*cos(D)} \; - \; \frac{sin(D)}{cos(H)*cos(D)}sin(L)$$

define constants A and B appropriately and the equation becomes:

$$cos(L) = A - B sin(L)$$

You could square both sides at this point, change cos2(L) to 1 - sin2(L) and solve the quadratic for sin(L).

RTW69

Or Sin(L)=[Sin(a)-Cos(L)*Cos(H)*Cos(D)]/Sin(D)

gneill

Mentor
Or Sin(L)=[Sin(a)-Cos(L)*Cos(H)*Cos(D)]/Sin(D)
How would that help? You've got a cos(L) on the RHS. You're trying to solve for L (eventually).

RTW69

You are correct, I should have read the problem statement a little closer.

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