Math Problem Statement

in triangle PQR, the measure of angle P is (3x+50, the measure of angle Q is (2x+9) and the measure of angle R is (4y+5) if side QR is extended through point R to point S, and the meauser of angle PRS is (x+y) what is the value of x+y

Solution

We are given a triangle PQRPQR with the following information:

  1. P=3x+50\angle P = 3x + 50
  2. Q=2x+9\angle Q = 2x + 9
  3. R=4y+5\angle R = 4y + 5
  4. When side QRQR is extended through point RR to point SS, the measure of PRS=x+y\angle PRS = x + y.

We need to determine the value of x+yx + y.

Step 1: Sum of angles in triangle PQRPQR

The sum of the interior angles of any triangle is 180180^\circ. Therefore, we can write the equation for the sum of the angles in triangle PQRPQR as:

(3x+50)+(2x+9)+(4y+5)=180(3x + 50) + (2x + 9) + (4y + 5) = 180

Simplify the left-hand side:

5x+4y+64=1805x + 4y + 64 = 180

Now, subtract 64 from both sides:

5x+4y=116(Equation 1)5x + 4y = 116 \quad \text{(Equation 1)}

Step 2: Exterior angle relationship

The exterior angle PRS\angle PRS is equal to the sum of the two non-adjacent interior angles of the triangle. Therefore, we have:

PRS=P+Q\angle PRS = \angle P + \angle Q

Substitute the given expressions for P\angle P and Q\angle Q:

x+y=(3x+50)+(2x+9)x + y = (3x + 50) + (2x + 9)

Simplify the right-hand side:

x+y=5x+59(Equation 2)x + y = 5x + 59 \quad \text{(Equation 2)}

Step 3: Solving the system of equations

Now, we have two equations:

  1. 5x+4y=1165x + 4y = 116
  2. x+y=5x+59x + y = 5x + 59

Solve Equation 2 for yy:

y=5x+59xy = 5x + 59 - x y=4x+59y = 4x + 59

Step 4: Substituting into Equation 1

Substitute y=4x+59y = 4x + 59 into Equation 1:

5x+4(4x+59)=1165x + 4(4x + 59) = 116

Simplify:

5x+16x+236=1165x + 16x + 236 = 116 21x+236=11621x + 236 = 116

Subtract 236 from both sides:

21x=12021x = -120

Solve for xx:

x=12021=407x = -\frac{120}{21} = -\frac{40}{7}

Step 5: Solving for yy

Substitute x=407x = -\frac{40}{7} into y=4x+59y = 4x + 59:

y=4(407)+59y = 4\left(-\frac{40}{7}\right) + 59 y=1607+59y = -\frac{160}{7} + 59 y=1607+4137=2537y = -\frac{160}{7} + \frac{413}{7} = \frac{253}{7}

Step 6: Finding x+yx + y

Now, calculate x+yx + y:

x+y=407+2537=2137x + y = -\frac{40}{7} + \frac{253}{7} = \frac{213}{7}

Thus, the value of x+yx + y is 2137\frac{213}{7} or approximately 30.43.

Let me know if you'd like further clarification or have any questions!


Here are 5 related questions to deepen your understanding:

  1. How do you apply the exterior angle theorem in different types of triangles?
  2. What is the importance of understanding angle relationships in polygons?
  3. How would the solution change if the angles were given in terms of other variables?
  4. Can you solve a triangle with only two angles given?
  5. How does the sum of angles change in non-Euclidean geometry?

Tip: Always double-check your equation simplifications to ensure the accuracy of your solution.

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Angle Sum in Triangles
Exterior Angle Theorem

Formulas

Sum of angles in a triangle: \( \angle P + \angle Q + \angle R = 180^\circ \)
Exterior angle theorem: \( \angle PRS = \angle P + \angle Q \)

Theorems

Angle sum theorem in a triangle
Exterior angle theorem

Suitable Grade Level

Grades 9-11