1. **Problem statement:** Find the angles marked with letters given that CAB is always a straight line. 2. **Part (a):** Equation given: $4x + 3x + 2x = 180$ This represents the sum of angles on a straight line, which is always $180^\circ$. 3. **Solve for $x$: ** Combine like terms: $$4x + 3x + 2x = 9x$$ So, $$9x = 180$$ Divide both sides by 9: $$\cancel{9}x = \frac{180}{\cancel{9}}$$ $$x = 20$$ 4. **Part (b) top-left triangle:** Given two angles inside the triangle: $30^\circ$ and $40^\circ$. Sum of angles in a triangle is $180^\circ$. Let the third angle inside the triangle be $z$. $$30 + 40 + z = 180$$ $$z = 180 - 70 = 110$$ Since $y$ is on the straight line AB, and angles on a straight line sum to $180^\circ$: $$y + z = 180$$ $$y = 180 - 110 = 70$$ 5. **Part (b) bottom-left obtuse angle:** Given an obtuse angle of $146^\circ$ on a straight horizontal line. Angles $a$, $b$, and $c$ are around a point near the line. Since $a$ is repeated twice and adjacent, and the total straight line is $180^\circ$: $$a + 146 = 180$$ $$a = 180 - 146 = 34$$ The angles $a$, $b$, and $c$ around a point sum to $360^\circ$: $$a + a + b + c = 360$$ Substitute $a=34$: $$34 + 34 + b + c = 360$$ $$68 + b + c = 360$$ $$b + c = 360 - 68 = 292$$ **Final answers:** - (a) $x = 20^\circ$ - (b) top-left: $y = 70^\circ$ - (b) bottom-left: $a = 34^\circ$, and $b + c = 292^\circ$ (cannot determine $b$ and $c$ individually with given info)