Literal Equation Isolate d

[ahohohoy]

Hello, im trying to make equation from a Law of Cossine. I have one triangle with side a, b and m and have all the values and i want to make another triangle using one vector that the direction is making that line c. The new triangle will have side c and d. So the law is a + b = c + d. The only thing that i could think of is Law of cosine, since I know the angle D is and the length of m. So could some one help me to isolate the length of d so i could find c and multiply that with the normalize vector so i could make the new triangle. Im stuck when trying to issolate the formula for d. So i make c = a + b - d and i make a new variable for a + b = s. So what i got is the black background image where i replace c with (s - d). Thanks.

 

[Steven G]

Any chance you explain why a+b=c+d as I do not see this being true? There are three angles at point D! When you refer to <D which one are you talking about?

Actually based on your own drawing a = sqrt(5), b = 2*sqrt(2), c= sqrt(2) and d= sqrt(5)

You need to explain how you are making this 2nd triangle!

 

[Dr.Peterson]

Hello, im trying to make equation from a Law of Cosine. I have one triangle with side a, b and m and have all the values and i want to make another triangle using one vector that the direction is making that line c. The new triangle will have side c and d. So the law is a + b = c + d. The only thing that i could think of is Law of cosine, since I know the angle D is and the length of m. So could some one help me to isolate the length of d so i could find c and multiply that with the normalize vector so i could make the new triangle. Im stuck when trying to isolate the formula for d. So i make c = a + b - d and i make a new variable for a + b = s. So what i got is the black background image where i replace c with (s - d). Thanks.

If I'm reading this correctly,



the LHS should be 2sd. Now gather terms in d by subtracting 2md cos(D) from both sides, factor out d, and divide

This assumes you are right that c+d=s=a+b, and m, s, and D are known, so that your equation is appropriate.

 

[ahohohoy]

Any chance you explain why a+b=c+d as I do not see this being true? There are three angles at point D! When you refer to <D which one are you talking about?

Actually based on your own drawing a = sqrt(5), b = 2*sqrt(2), c= sqrt(2) and d= sqrt(5)

You need to explain how you are making this 2nd triangle!

thats what i want to make. So im making a leg using this, and the rule for the leg to works is a+b should be equal to c+d, and i need to find what is the value of d, so i could do c = a +b - d, and find the c value to build the second triangle. D is the angle between m and c, thats the angle that I know currently.

 

[ahohohoy]

If I'm reading this correctly,

View attachment 31722

the LHS should be 2sd. Now gather terms in d by subtracting 2md cos(D) from both sides, factor out d, and divide

This assumes you are right that c+d=s=a+b, and m, s, and D are known, so that your equation is appropriate.

Oh yeah, it should be 2sd, thanks.yeah, for the leg to move correctly (im making a leg system), a + b should be equal c + d to works. Okay is it what you mean?? Yeah im trying to isolate d = ... , so i could get the c = a + b - d, and could make the second triangle that has the same length along the normal of that long vector of c.

 

[ahohohoy]

If I'm reading this correctly,

View attachment 31722

the LHS should be 2sd. Now gather terms in d by subtracting 2md cos(D) from both sides, factor out d, and divide

This assumes you are right that c+d=s=a+b, and m, s, and D are known, so that your equation is appropriate.

Okay it works!!! Thank you so much dr.peterson, i have to learn more about algebra and using factorization.