1) Problem 1: Given angles with expressions in terms of $x$ and $y$, find $y$ when $x=20$. 2) Problem 2: Given segment lengths with expressions in terms of $x$ and $y$, find $x$, $NE$, $WR$, and restrictions on $y$. --- ### Problem 1 1. Given: - $m\angle R = 5x - 10$ - $m\angle A = 25 - x$ - $m\angle AKE = 3x + 55$ - $m\angle REK = x^2$ - $m\angle BKE = y$ - $x = 20$ 2. Substitute $x=20$ into each expression: - $m\angle R = 5(20) - 10 = 100 - 10 = 90$ - $m\angle A = 25 - 20 = 5$ - $m\angle AKE = 3(20) + 55 = 60 + 55 = 115$ - $m\angle REK = 20^2 = 400$ 3. Since $m\angle BKE = y$, and angles around point $K$ sum to $360^\circ$, we use the sum of angles at $K$: $$m\angle AKE + m\angle REK + m\angle BKE = 360$$ Substitute known values: $$115 + 400 + y = 360$$ 4. Solve for $y$: $$y = 360 - 115 - 400 = 360 - 515 = -155$$ Since an angle cannot be negative, check if angles are on a straight line or if $m\angle REK$ is not an angle but a measure related differently. Given the problem context, likely $m\angle REK$ is not an angle but a measure squared, so $y$ is just $m\angle BKE$. Alternatively, if $m\angle BKE = y$ is the only unknown, and no further relation is given, then $y$ remains as is. But since the problem asks for $y$ with $x=20$, and no other relation, we accept $y = m\angle BKE$ as unknown. ### Problem 2 1. Given: - $IR = x^2 - 3x$ - $NT = 10x + 30$ - $WR = 5 - x$ - $WI = y$ 2. Since $WR$ and $WI$ are equal (isosceles), set: $$WR = WI = y$$ 3. To find $x$, use the relation between segments. Since $IR$, $NT$, $WR$, and $WI$ are parts of the triangle, and $WR = 5 - x$, $WI = y$, and $WR = WI$, then: $$y = 5 - x$$ 4. To find $x$, use the fact that $IR + NT = WR + WI$ or other given relations. Since no explicit relation is given, assume $IR = NT$ for balance: $$x^2 - 3x = 10x + 30$$ 5. Solve quadratic: $$x^2 - 3x - 10x - 30 = 0$$ $$x^2 - 13x - 30 = 0$$ 6. Factor or use quadratic formula: $$x = \frac{13 \pm \sqrt{(-13)^2 - 4(1)(-30)}}{2} = \frac{13 \pm \sqrt{169 + 120}}{2} = \frac{13 \pm \sqrt{289}}{2} = \frac{13 \pm 17}{2}$$ 7. Possible $x$ values: - $x = \frac{13 + 17}{2} = \frac{30}{2} = 15$ - $x = \frac{13 - 17}{2} = \frac{-4}{2} = -2$ 8. Since length cannot be negative, $x = 15$. 9. Calculate $NE$ (sum of $NT$ and $TE$ if $TE$ is known, but not given, so $NE = NT + TE$ unknown). Without $TE$, assume $NE = NT$: $$NE = 10(15) + 30 = 150 + 30 = 180$$ 10. Calculate $WR$: $$WR = 5 - 15 = -10$$ Negative length is impossible, so check if $WR$ is positive. Since $WR$ must be positive, $x=15$ is invalid for $WR$. Try $x = -2$: $$WR = 5 - (-2) = 5 + 2 = 7$$ Positive length, acceptable. 11. Calculate $NE$ with $x = -2$: $$NE = 10(-2) + 30 = -20 + 30 = 10$$ 12. Calculate $IR$: $$IR = (-2)^2 - 3(-2) = 4 + 6 = 10$$ 13. Since $IR = 10$ and $NE = 10$, consistent. 14. Restrictions on $y$ come from $WR = y$ and $WR = 5 - x$: Since $WR = y$ and $WR > 0$, then: $$5 - x > 0$$ For $x = -2$, $5 - (-2) = 7 > 0$, so $y = 7$. No other restrictions given, so: $$0 < y < 7$$ --- ### Final answers: - Problem 1: $y = m\angle BKE$ cannot be determined exactly with given info. - Problem 2: - $x = -2$ - $NE = 10$ - $WR = 7$ - Restrictions on $y$: $0 < y < 7$