1. **Problem 1:** Convert $(x - 1)^\circ = \frac{3\pi}{20}$ radians to degrees and solve for $x$. 2. Recall the conversion formula between radians and degrees: $$\text{degrees} = \text{radians} \times \frac{180}{\pi}$$ 3. Convert the right side to degrees: $$\frac{3\pi}{20} \times \frac{180}{\pi} = \frac{3 \times 180}{20} = \frac{540}{20} = 27^\circ$$ 4. Set up the equation: $$(x - 1)^\circ = 27^\circ$$ 5. Solve for $x$: $$x - 1 = 27$$ $$x = 27 + 1 = 28$$ --- 1. **Problem 2:** Given two angles on a straight line: $(x + 40)^\circ$ and $(20 - x)^\circ$, find $x$. 2. Angles on a straight line sum to $180^\circ$, so: $$ (x + 40) + (20 - x) = 180 $$ 3. Simplify the left side: $$ x + 40 + 20 - x = 60 $$ 4. So: $$ 60 = 180 $$ This is a contradiction, so check carefully. The problem likely asks to determine $x$ such that the sum equals one of the given options. Since the sum is always 60, none of the options match 180. Possibly the problem wants the value of the sum or $x$ for which the sum equals one of the options. Alternatively, if the problem is to find $x$ such that the sum equals 180, then: $$ (x + 40) + (20 - x) = 180 $$ $$ 60 = 180 $$ No solution. Therefore, the problem might be to find $x$ such that the two angles are equal: $$ x + 40 = 20 - x $$ $$ 2x = -20 $$ $$ x = -10 $$ Check the options: none match -10. Since the problem is ambiguous, we proceed with the sum equals 180 assumption and find no solution. --- 1. **Problem 3:** Given three angles: $50^\circ - 2x$, $10^\circ + x$, and $20^\circ + x$, find $x$. 2. These three angles likely form a triangle, so their sum is $180^\circ$: $$ (50 - 2x) + (10 + x) + (20 + x) = 180 $$ 3. Simplify: $$ 50 - 2x + 10 + x + 20 + x = 180 $$ $$ (50 + 10 + 20) + (-2x + x + x) = 180 $$ $$ 80 + 0 = 180 $$ 4. This simplifies to: $$ 80 = 180 $$ Which is false, so check if the problem is to find $x$ such that two angles are equal or some other condition. If the problem is to find $x$ such that $10 + x = 20 + x$, then: $$ 10 + x = 20 + x $$ $$ 10 = 20 $$ False. If the problem is to find $x$ such that $50 - 2x = 10 + x$, then: $$ 50 - 2x = 10 + x $$ $$ 50 - 10 = 2x + x $$ $$ 40 = 3x $$ $$ x = \frac{40}{3} \approx 13.33 $$ No matching option. Alternatively, if the problem is to find $x$ such that $50 - 2x = 20 + x$, then: $$ 50 - 2x = 20 + x $$ $$ 50 - 20 = 2x + x $$ $$ 30 = 3x $$ $$ x = 10 $$ Closest option is a) 10°. **Final answers:** - Problem 1: $x = 28$ - Problem 2: No solution for $x$ making sum 180°; sum is always 60° - Problem 3: $x = 10$ (option a)