MHB Equation to find an unknown point on a graph

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To find the location of point 'E' on a graph given known values for points 'B' and 'C' and angle 'D', coordinate geometry can be utilized. The equations for the horizontal and sloping lines are established, allowing for the calculation of 'E's coordinates as (b+c*cos(δ)/sin(δ), c). The distance from 'A' to 'E' can be determined using the formula √(b²+c²+2bc*cos(δ))/sin(δ), and the angle 'θ' that line 'AE' makes with the horizontal is found using tan(θ) = c*sin(δ)/(b+c*cos(δ)). This method provides a general solution for locating 'E' regardless of the values for 'B', 'C', and 'D'.
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Hi All,

I need help to find an equation that will give me the location of 'E' in the image below.

'A' is the starting point on a graph.

All values from 'B' to 'D' are known values.

Currently, 'B' = 50, 'C' = 20, & 'D' = 135° - but I need an equation that will find 'E' no matter what these values are.

View attachment 7931

Any help would be greatly appreciated!

- Jazoar
 

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Jazoar said:
Hi All,

I need help to find an equation that will give me the location of 'E' in the image below.

'A' is the starting point on a graph.

All values from 'B' to 'D' are known values.

Currently, 'B' = 50, 'C' = 20, & 'D' = 135° - but I need an equation that will find 'E' no matter what these values are.
Any help would be greatly appreciated!

- Jazoar
Hi Jazoar, and welcome to MHB!

I found your diagram difficult to follow, because the letters A, B, C, D and E are being used to describe different things. If I understand it correctly, A and E are points, B and C are distances, and D is an angle. So I shall change the notation, keeping $A$ and $E$ as points but writing the distances as $b$ and $c$, and the angle as $\delta.$

I think the easiest way to solve the problem is to use coordinate geometry, taking $A$ as the origin, and the horizontal blue line as the $x$-axis. The horizontal red line then has the equation $y=c$, and the sloping red line has equation $-y\cos\delta + x\sin\delta = b.$ Those two lines meet at the point $E$, whose coordinates are $\left(\frac{b+c\cos\delta}{\sin\delta},c\right).$

From that, you can calculate that the distance from $A$ to $E$ is $$\frac{\sqrt{b^2+c^2 + 2bc\cos\delta}}{\sin\delta},$$ and if the line $AE$ makes an angle $\theta$ with the horizontal then $\theta$ is given by $$\tan\theta = \frac{c\sin\delta}{b+c\cos\delta}.$$ In the case when $b=50$, $c=20$ and $\delta = 135^\circ$, those formulas give $AE \approx 54.5$ and $\theta \approx 21.5^\circ.$
 
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