Answer:
[tex]BA = 6.0[/tex]
[tex]BC = 13.4[/tex]
Step-by-step explanation:
Given
The attached triangle
Solving (a) BA
Considering the tangent of angle C, we have:
[tex]tan\ C = \frac{BA}{AC}\\[/tex]
This gives:
[tex]tan\ 25 = \frac{BA}{12}[/tex]
Make BA the subject
[tex]BA = 12 * tan\ 25[/tex]
[tex]BA = 12 * 0.4663[/tex]
[tex]BA = 6.0[/tex] --- approximated
Solving (a) BC
Using Pythagoras theorem
[tex]BC^2 = BA^2 + AC^2[/tex]
[tex]BC^2 = 6.0^2 + 12^2[/tex]
[tex]BC^2 = 180[/tex]
Square root of both sides
[tex]BC = \sqrt{180[/tex]
[tex]BC = 13.4[/tex]
