Home computickets Search Points on the map by two coordinates and distances to it

# Search Points on the map by two coordinates and distances to it

Author

Date

Category

figured out how to search the distance between two points on the map, but I can’t understand how to solve the opposite task.

There are three points on the map with known coordinates A (0,0) B (0.90) C (90.0) and are known to the distance from them to point d, the question is how to find the coordinates of this point. As I understood, it is enough to solve the spherical triangle ABD, and the distance with use only to determine the pole. But with the calculation of this, problems arose.

UPD led to the task to solve the ABC triangle, following the example https://www.math10.com/ru/vysshaya-matematika/sfericheskii-treugolnik/sfericheskii-treugolnik.html

``````a (0,0) B (0.90) C (? ,?)
AC = 8000; AB = 10007; Bc = 4000;
a = 1.255; B = 1.571; C = 0.628;
COS A = (COS A - COS B * COS C) / SIN B * SIN C
(0.370-0 * 0.809) /1*.587=0.630
``````

I can not find which corner meets this cosine, and how can you get bearing to calculate the coordinates? Or did I do something wrong?

You know the angular lengths of the sides (distances divided by the land radius)

Apply spherical cosine theorem , To find the inner corner at a

point

``````cos a = cos b ⋅ cos c + sin b ⋅ sin c ⋅ cos a
``````

from here (using D, and not with, because triangle ABD)

``````a = arccos ((COS A - COS B ⋅ COS D) / (SIN B ⋅ SIN D))
for ad = 8000; AB = 10007; Bd = 4000;
opposing top and side
a = bd / 6371 = 0.6278
B = AD / 6371 = 1.255
D = AB / 6371 = 1.571
A = ArcCOS ((COS A - COS B ⋅ COS D) / (SIN B ⋅ SIN D)) =
ArcCOS ((0.8093 - 0.3106 ⋅ 0) / (0.9505 ⋅ 1)) =
ArcCOS (0.851) = 0.552 radians = 31.68 degrees
``````

This is azimuth from point A (in this case, azimuth is immediately, since A and B at the equator)

then use distance and azimuth (a) to determine the coordinates D, as shown Here in the section `Destination Point Given Distance and Bearing from Start Point `

``````φ2 = asin (sin φ1 ⋅ cos δ + cos φ1 ⋅ sin δ ⋅ cos θ)
λ2 = λ1 + atan2 (sin θ ⋅ sin δ ⋅ cos φ1, cos δ - sin φ1 ⋅ sin φ2)
Where φ is Latitude, λ is Longitude, θ is the Bearing (Clockwise Fråm
North), Δ is the angular distance d / r; D Being The Distance Travelled,
Lat = arcsin (sin φ1 ⋅ cos δ + cos φ1 ⋅ sin δ ⋅ cos θ) =
Arcsin (0 * COS B + 1 * SIN B * SIN (0.552)) =
Arcsin (0.9505 * 0.524) =
Arcsin (0.4984) = 29.9 degrees (SS or Yush)
And then for Lon
``````

Selecting a hemisphere for D – Yes, as already said, away from the pole

Why spend time searching for the correct question and then entering your answer when you can find it in a second? That's what CompuTicket is all about! Here you'll find thousands of questions and answers from hundreds of computer languages.