# Trig. identities method problem

1. Oct 19, 2006

### DAZZLED

Question
I've been struggling with the followings equations (I know that can be solved by Trig. identities method) and I tried almost everything but I can't get them right.
the equations are the following:
-Tab*cos(alpha)+Tbc*cos(beta) = 0 (eq.1)
Tab*sin(alpha)+Tbc*sin(beta)=15
We know the angular values for alpha and beta, and also I reviewed the answers on my workbook and they are; Tab=9.67 and Tbc=12.334, I would like if someone can show me how to solve those identities step by a step.
Those equations are part of a problem in mechanics of materials, in which two wires are supporting a weigth (15) both in different known angles (that's why the equations of equilibrium yield those two precedent equations) but I can't get the correct results for each tension in the wires, I know they are trig. identities and something I am doing is not coming well enough in the algebraic procedure.
Thanks

2. Oct 19, 2006

### Naumaan

According to the question
for my convenience let alpha=A and beta=B
there are two equations
Tbc*cosB - Tab*cosA = 0 ------i
Tab*sinA+Tbc*sinB=15-------ii
From i
Tbc*cosB = Tab*cosA
=> Tbc=Tab*cosA/cosB dividing by cosB both sides
Placing this value in ii
=> Tab*sinA + (Tab*cosA/cosB)*sinB = 15
=> Tab*(sinA + cosAsinB/cosB) = 15
=> Tab*(sinAcosB + cosAsinB)/cosB = 15
=> Tab*sin(A + B) / cosB = 15 As sin(A+B)=(sinA*cosB)+(cosA*sinB)
=> Tab = 15*cosB/sin(A + B)
Placing this value in
Tbc = Tab * cosA / cosB
now u can calculate the values

3. Oct 19, 2006

### HallsofIvy

Staff Emeritus
Naumaan gave a perfectly good answer but I would do it a slightly different way.
Multiply the first equation by $T_{ab}cos(\alpha)+T_{bc}cos(\beta)$ to get $T_{bc} cos^2(\alpha)-T_{ab}cos^2(\beta)= 0$.
Multiply the second equation by $T_{ab} sin(\alpha)- T_{bc}sin(\beta)$ to get $T_{ab}^2 sin^2(\alpha)- T_{bc}^2 sin^2(\beta)= 15T_{ab}sin(\alpha)- 15T_{bc}sin(\beta$ and add the two equations. since $sin^2(x)+ cos^2(x)= 1$, that gives $T_{ab}^2 - T_{bc}^2= 15T_{ab}sin(\alpha)- 15T_{bc}sin(\beta$ which involves only sines of $\alpha$ and $\beta$.

Use $T_{ab}sin(\alpha)+ T_{bc}sin(\beta)= 15$ to replace either $sin(\alpha)$ or $sin(\beta)$ by the other and solve the resulting equation.