QUESTION IMAGE
Question
honors extension: solve for the variables. 1) (4z + 8)° x° 106° 2y° 2) (9x + 12)° 3x° (4y - 10)° 3) 3x° 2y° 4y° (5x - 20)° 4) 90° (15x + 30)° (3y + 18)° 10x° 5) 4x° (3y + 5)° 65° 52° 6) 2y° z° 2x° 90° x° 7) 120° (2x + y)° (2x - y)° 140° 8) 50° x° 100° 9) 30° 5y° 2x° (x - y)° a b
Step1: Use vertical - angle property for problem 1
Vertical angles are equal. So $x = 106$. Also, since $2y$ and $106$ are supplementary (linear - pair), $2y+106 = 180$. Then $2y=180 - 106=74$, and $y = 37$. And $4z + 8+106 = 180$ (linear - pair), so $4z+8=74$, $4z = 66$, $z=\frac{66}{4}=16.5$.
Step2: Use vertical - angle property for problem 2
Vertical angles are equal. So $9x + 12=3x+180$ (the sum of angles around a point). $9x-3x=180 - 12$, $6x = 168$, $x = 28$. Also, $4y-10 = 3x$, substituting $x = 28$, we get $4y-10=3\times28$, $4y-10 = 84$, $4y=94$, $y = 23.5$.
Step3: Use angle - sum property of a triangle for problem 3
In the larger triangle, the sum of interior angles is $180^{\circ}$. So $2y + 4y+(5x - 20)=180$, $6y+5x-20 = 180$, $6y+5x=200$. Also, in the smaller triangle with angle $3x$, we know that $2y+3x+(180-(5x - 20))=180$, $2y+3x - 5x + 20=0$, $2y-2x=-20$, $y - x=-10$, $y=x - 10$. Substitute $y=x - 10$ into $6y+5x=200$: $6(x - 10)+5x=200$, $6x-60+5x=200$, $11x=260$, $x=\frac{260}{11}\approx23.64$, $y=\frac{260}{11}-10=\frac{260 - 110}{11}=\frac{150}{11}\approx13.64$.
Step4: Use angle - sum property of a quadrilateral for problem 4
The sum of interior angles of a quadrilateral is $360^{\circ}$. So $90+(15x + 30)+10x+(3y + 18)=360$, $25x+3y+138 = 360$, $25x+3y=222$. Also, we can use the fact that the figure may have parallel - side properties (if applicable) to get more relationships. But if we assume it's a general quadrilateral, we have one equation with two variables. If we assume some special - case (e.g., trapezoid with right - angles), we may need more information.
Step5: Use angle - sum property of a triangle for problem 5
In the right - triangle, $4x+(3y + 5)+90=180$, $4x+3y=85$. Also, in the other triangle with angles $65^{\circ}$ and $52^{\circ}$, we know that the third angle is $180-(65 + 52)=63^{\circ}$. But this doesn't give us a direct relation to solve for $x$ and $y$ without more information.
Step6: Use angle - sum property of a triangle and linear - pair property for problem 6
We know that $2x+x + 90=180$ (angle - sum of a triangle), $3x=90$, $x = 30$. Also, $2y+z=180$ (linear - pair) and we can use other angle relationships in the figure. But with the given information, if we just focus on the triangle with $2x$, $x$ and $90^{\circ}$, we found $x = 30$.
Step7: Use angle - sum property of a quadrilateral for problem 7
The sum of interior angles of a quadrilateral is $360^{\circ}$. So $120+(2x + y)+(2x - y)+140=360$, $4x+260 = 360$, $4x=100$, $x = 25$.
Step8: Use angle - sum property of a polygon (in this case, a non - regular hexagon formed by the parallel lines)
We know that the sum of exterior angles of a polygon is $360^{\circ}$. But we can also use the fact that the angles are related by parallel - line properties. If we consider the angles $50^{\circ}$, $100^{\circ}$ and $x$, we know that $x=130$ (by the property of angles formed by parallel lines and a transversal).
Step9: Use vertical - angle and angle - sum property of a triangle for problem 9
Vertical angles: $5y = 30$, $y = 6$. In the triangle with angles $2x$, $(x - y)$ and the angle related to the vertical - angle, using the angle - sum property of a triangle $2x+(x - y)+30=180$. Substitute $y = 6$ into it: $2x+(x - 6)+30=180$, $3x+24 = 180$, $3x=156$, $x = 52$.
- For problem 1:
- $x = 106$, $y = 37$, $z = 16.5$
- For problem 2:
- $x = 28$, $y = 23.5$
- For problem 3:
- $x=\frac{260}{11}\approx23.64$, $y=\frac{150}{11}\approx13.64$
- For problem 4:
- Need more information to solve uniquely for $x$…
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Step1: Use vertical - angle property for problem 1
Vertical angles are equal. So $x = 106$. Also, since $2y$ and $106$ are supplementary (linear - pair), $2y+106 = 180$. Then $2y=180 - 106=74$, and $y = 37$. And $4z + 8+106 = 180$ (linear - pair), so $4z+8=74$, $4z = 66$, $z=\frac{66}{4}=16.5$.
Step2: Use vertical - angle property for problem 2
Vertical angles are equal. So $9x + 12=3x+180$ (the sum of angles around a point). $9x-3x=180 - 12$, $6x = 168$, $x = 28$. Also, $4y-10 = 3x$, substituting $x = 28$, we get $4y-10=3\times28$, $4y-10 = 84$, $4y=94$, $y = 23.5$.
Step3: Use angle - sum property of a triangle for problem 3
In the larger triangle, the sum of interior angles is $180^{\circ}$. So $2y + 4y+(5x - 20)=180$, $6y+5x-20 = 180$, $6y+5x=200$. Also, in the smaller triangle with angle $3x$, we know that $2y+3x+(180-(5x - 20))=180$, $2y+3x - 5x + 20=0$, $2y-2x=-20$, $y - x=-10$, $y=x - 10$. Substitute $y=x - 10$ into $6y+5x=200$: $6(x - 10)+5x=200$, $6x-60+5x=200$, $11x=260$, $x=\frac{260}{11}\approx23.64$, $y=\frac{260}{11}-10=\frac{260 - 110}{11}=\frac{150}{11}\approx13.64$.
Step4: Use angle - sum property of a quadrilateral for problem 4
The sum of interior angles of a quadrilateral is $360^{\circ}$. So $90+(15x + 30)+10x+(3y + 18)=360$, $25x+3y+138 = 360$, $25x+3y=222$. Also, we can use the fact that the figure may have parallel - side properties (if applicable) to get more relationships. But if we assume it's a general quadrilateral, we have one equation with two variables. If we assume some special - case (e.g., trapezoid with right - angles), we may need more information.
Step5: Use angle - sum property of a triangle for problem 5
In the right - triangle, $4x+(3y + 5)+90=180$, $4x+3y=85$. Also, in the other triangle with angles $65^{\circ}$ and $52^{\circ}$, we know that the third angle is $180-(65 + 52)=63^{\circ}$. But this doesn't give us a direct relation to solve for $x$ and $y$ without more information.
Step6: Use angle - sum property of a triangle and linear - pair property for problem 6
We know that $2x+x + 90=180$ (angle - sum of a triangle), $3x=90$, $x = 30$. Also, $2y+z=180$ (linear - pair) and we can use other angle relationships in the figure. But with the given information, if we just focus on the triangle with $2x$, $x$ and $90^{\circ}$, we found $x = 30$.
Step7: Use angle - sum property of a quadrilateral for problem 7
The sum of interior angles of a quadrilateral is $360^{\circ}$. So $120+(2x + y)+(2x - y)+140=360$, $4x+260 = 360$, $4x=100$, $x = 25$.
Step8: Use angle - sum property of a polygon (in this case, a non - regular hexagon formed by the parallel lines)
We know that the sum of exterior angles of a polygon is $360^{\circ}$. But we can also use the fact that the angles are related by parallel - line properties. If we consider the angles $50^{\circ}$, $100^{\circ}$ and $x$, we know that $x=130$ (by the property of angles formed by parallel lines and a transversal).
Step9: Use vertical - angle and angle - sum property of a triangle for problem 9
Vertical angles: $5y = 30$, $y = 6$. In the triangle with angles $2x$, $(x - y)$ and the angle related to the vertical - angle, using the angle - sum property of a triangle $2x+(x - y)+30=180$. Substitute $y = 6$ into it: $2x+(x - 6)+30=180$, $3x+24 = 180$, $3x=156$, $x = 52$.
- For problem 1:
- $x = 106$, $y = 37$, $z = 16.5$
- For problem 2:
- $x = 28$, $y = 23.5$
- For problem 3:
- $x=\frac{260}{11}\approx23.64$, $y=\frac{150}{11}\approx13.64$
- For problem 4:
- Need more information to solve uniquely for $x$ and $y$
- For problem 5:
- Need more information to solve uniquely for $x$ and $y$
- For problem 6:
- $x = 30$
- For problem 7:
- $x = 25$
- For problem 8:
- $x = 130$
- For problem 9:
- $x = 52$, $y = 6$