I have to solve a system of nonlinear equations using jacobian but I'm not sure how to solve for the solutions. I remember one of my friends doing $Ax = B$; where jacobian matrix was $A$, but im not sure what was $B$.
$10\cos(t_3)-9\cos(t_4) = 8 - 3.5\cos(15)$
$10\sin(t_3)-9\sin(t_4) = -3.5\sin(15)$
I have to solve for $t_3$ and $t_4$. And it tells me to set it in terms of $t_2$ (which is $15$ in the problem). Answers were $t_3 = 53$ degrees and $t_4 = 81$ degrees
Hint
Let me start very basic : you want to solve $F(x,y)=0$ and $G(x,y)=0$ starting from an initial guess $(x_0,y_)$.
Expand each equation as a first order Taylor series. So $$F(x_0,y_0)+F'_x (x_0,y_0)(x-x_0)+F'_y (x_0,y_0)(y-y_0)=0$$ $$G(x_0,y_0)+G'_x (x_0,y_0)(x-x_0)+G'_y (x_0,y_0)(y-y_0)=0$$ or $$F'_x (x_0,y_0) \Delta X+F'_y (x_0,y_0)\Delta Y=-F(x_0,y_0)$$ $$G'_x (x_0,y_0)\Delta X+G'_y (x_0,y_0)\Delta Y=-G(x_0,y_0)$$ I suppose you see $A$ and $B$.
But since the problem is not linear, you must restart the calculation using $x_0=x_0+\Delta X$, $y_0=y_0+\Delta Y$ untill they do not change for the required accuracy.
I am sure that you can take from here and solve your specific problem.
Applying the method, you should arrive to $t_3=52.9807$ and $t_4=81.0408$ (I suggest you do all the work using radians to make you life simpler). There is another solution.