Respuesta :
Answer:
[tex]Side\ B = 6.0[/tex]
[tex]\alpha = 56.3[/tex]
[tex]\theta = 93.7[/tex]
Step-by-step explanation:
Given
Let the three sides be represented with A, B, C
Let the angles be represented with [tex]\alpha, \beta, \theta[/tex]
[See Attachment for Triangle]
[tex]A = 10cm[/tex]
[tex]C = 12cm[/tex]
[tex]\beta = 30[/tex]
What the question is to calculate the third length (Side B) and the other 2 angles ([tex]\alpha\ and\ \theta[/tex])
Solving for Side B;
When two angles of a triangle are known, the third side is calculated as thus;
[tex]B^2 = A^2 + C^2 - 2ABCos\beta[/tex]
Substitute: [tex]A = 10[/tex], [tex]C =12[/tex]; [tex]\beta = 30[/tex]
[tex]B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30[/tex]
[tex]B^2 = 100 + 144 - 240*0.86602540378[/tex]
[tex]B^2 = 100 + 144 - 207.846096907[/tex]
[tex]B^2 = 36.153903093[/tex]
Take Square root of both sides
[tex]\sqrt{B^2} = \sqrt{36.153903093}[/tex]
[tex]B = \sqrt{36.153903093}[/tex]
[tex]B = 6.0128115797[/tex]
[tex]B = 6.0[/tex] (Approximated)
Calculating Angle [tex]\alpha[/tex]
[tex]A^2 = B^2 + C^2 - 2BCCos\alpha[/tex]
Substitute: [tex]A = 10[/tex], [tex]C =12[/tex]; [tex]B = 6[/tex]
[tex]10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha[/tex]
[tex]100 = 36 + 144 - 144 *Cos\alpha[/tex]
[tex]100 = 36 + 144 - 144 *Cos\alpha[/tex]
[tex]100 = 180 - 144 *Cos\alpha[/tex]
Subtract 180 from both sides
[tex]100 - 180 = 180 - 180 - 144 *Cos\alpha[/tex]
[tex]-80 = - 144 *Cos\alpha[/tex]
Divide both sides by -144
[tex]\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}[/tex]
[tex]\frac{-80}{-144} = Cos\alpha[/tex]
[tex]0.5555556 = Cos\alpha[/tex]
Take arccos of both sides
[tex]Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)[/tex]
[tex]Cos^{-1}(0.5555556) = \alpha[/tex]
[tex]56.25098078 = \alpha[/tex]
[tex]\alpha = 56.3[/tex] (Approximated)
Calculating [tex]\theta[/tex]
Sum of angles in a triangle = 180
Hence;
[tex]\alpha + \beta + \theta = 180[/tex]
[tex]30 + 56.3 + \theta = 180[/tex]
[tex]86.3 + \theta = 180[/tex]
Make [tex]\theta[/tex] the subject of formula
[tex]\theta = 180 - 86.3[/tex]
[tex]\theta = 93.7[/tex]
