Triangle RTS is on a horizontal line. Line SR extends through point Q to form exterior angle TRQ. Angle RTS is [tex](25x)[/tex] degrees. Angle TSR is [tex](57 + x)[/tex] degrees. Exterior angle TRQ is [tex](45x)[/tex] degrees. Find the value of x.
Options: x = 2, x = 3, x = 33, x = 52.
Answer (2)
To solve for the value of x , we need to use the properties of angles in a triangle and the exterior angle theorem.
Understanding the Problem :
We have triangle RTS where:
∠ RTS = 25 x degrees.
∠ TSR = 57 + x degrees.
Exterior angle ∠ TRQ = 45 x degrees.
Use the Exterior Angle Theorem :
The Exterior Angle Theorem states that the exterior angle is equal to the sum of the two non-adjacent interior angles.
Set up the equation :
∠ TRQ = ∠ RTS + ∠ TSR
Substituting the given values: 45 x = 25 x + ( 57 + x )
Simplify: 45 x = 25 x + 57 + x 45 x = 26 x + 57
Solve for x by first moving terms involving x to one side: 45 x − 26 x = 57 19 x = 57
Divide both sides by 19: x = 19 57 x = 3
Conclusion :
The value of x is 3.
Therefore, the correct answer is x = 3 .
The value of x in the given triangle is 3, as determined by solving the equation derived from the Exterior Angle Theorem. The correct answer is therefore x = 3 .
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