# Angular intersection.... i think trig is used but im well confused pls pls pls help

• Aug 20th 2009, 03:38 AM
jane??
Angular intersection.... i think trig is used but im well confused pls pls pls help
Intersection.... how would u do this?
the following clock wise angles were observed:
at A from B to P the angle is 64d 42m 42s
at B from A to P the angle is 296d 15m 31s

the co-ordinates of A are 500E 500N
the co-ordinates of B are 703.501E 240.809N
Find the co-ordinates of P
• Aug 20th 2009, 06:02 AM
skeeter
Quote:

Originally Posted by jane??
Intersection.... how would u do this?
the following clock wise angles were observed:
at A from B to P the angle is 64d 42m 42s
at B from A to P the angle is 296d 15m 31s

the co-ordinates of A are 500E 500N
the co-ordinates of B are 703.501E 240.809N
Find the co-ordinates of P

are these three points on a plane surface or a sphere?
• Aug 20th 2009, 10:27 AM
jane??
plane surface?? it doesnt say in the question but thats wat i would presume
• Aug 21st 2009, 02:10 PM
aidan
Quote:

Originally Posted by jane??
Intersection.... how would u do this?
the following clock wise angles were observed:
at A from B to P the angle is 64d 42m 42s
at B from A to P the angle is 296d 15m 31s

the co-ordinates of A are 500E 500N
the co-ordinates of B are 703.501E 240.809N
Find the co-ordinates of P

See the attached image.

C is due north of point P &
C is due west of point A
D is due east of A
Points D, A, & C all have north coordinate of 500.000
---
The difference of the Northing coordinates from A to B:
( 240.809N - 500.000N ) = -259.191

The difference of the Easting coordinates from A to B:
( 703.501E - 500.000E ) = +203.501

The distance AB can be computed:
$\displaystyle \bar{AB} = \sqrt{ (-259.191)^2 + 203.501^2 } = 329.53396$

At this moment you should calculate the angle DAB
since it will be needed later:

$\displaystyle \angle DAB = arctan \left ( \dfrac{259.191}{203.501} \right ) = 51d 51m 47s$

[NOTE: We know that DAB is a negative angle, but that attribute will be ignored]
Angle CAP = 180d - ( 51d 51m 47s ) - ( 64d 42m 42s) = 63d 25m 31s <- NEED THIS!

Angle APC = 90d - ( 63d 25m 31s ) = 26d 34m 29s <-- do not need this but there it is.

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You are given the interior angle BAP = 64d 42m 42s
The interior angle at B is 360d - (296d 15m 31s) = 63d 44m 29m
The third angle at P is: 180deg minus angles A & B
::
180d - (64d 42m 42s) - (63d 44m 29m) = 51d 32m 49s

Using the Rule of Sines (or The Law of Sines, or Sine Law)

$\displaystyle \dfrac{ AB }{ \sin (51d 32m 49s) } = \dfrac{ AP }{ \sin (63d 44m 29m)}$

$\displaystyle \dfrac{ AB \times \sin (63d 44m 29m) }{ \sin (51d 32m 49s) } = AP$

We do not require the distance BP.

---
The northing coordinates of P:
North coord of A MINUS distance AP x sin(angle CAP)

$\displaystyle 500.000 - AP \cdot \sin(63d 25m 31s)$

---
The easting coordinates of P:
East coords of A MINUS distance AP x cos(angle CAP)

$\displaystyle 500.000 - AP \cdot \sin(63d 25m 31s)$

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Run this and determine the values.
I will do the math parts later as a check.
From the image it is clear that Point P is SOUTH & WEST of Point A.
If you do not have the visual aid, it is imperative to be cognizant of the signs of the computed values.
• Aug 21st 2009, 02:14 PM
jane??
thanks so much your a star :) :) i put up another question today can u have a look at it.... i wish i was good at maths lol