1. **State the problem:** We need to find the size of angle $t$ in an irregular pentagon with given angles: $129^\circ$, $117^\circ$, $40^\circ$, $40^\circ$, and $137^\circ$. There is also a right angle ($90^\circ$) indicated at the bottom-left vertex. 2. **Recall the sum of interior angles of a pentagon:** The sum of interior angles in any pentagon is given by the formula: $$\text{Sum} = (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$$ 3. **Sum the known angles:** Add all the known angles including the right angle: $$129^\circ + 117^\circ + 40^\circ + 40^\circ + 137^\circ + 90^\circ$$ 4. **Check the problem carefully:** The problem states the pentagon has five angles, but the right angle is indicated as a small square at the bottom-left vertex, which is one of the five vertices. So the right angle is one of the five angles, not an extra angle. The angles given are $129^\circ$, $117^\circ$, $40^\circ$, $40^\circ$, $137^\circ$, and $t$. The right angle is $90^\circ$ and corresponds to one of these angles, so we must identify which one. 5. **Identify the right angle:** The problem states the small square is at the bottom-left vertex, which corresponds to one of the $40^\circ$ angles or the $137^\circ$ angle? Since $40^\circ$ is too small for a right angle, and $137^\circ$ is obtuse, the right angle is likely the $90^\circ$ angle at the bottom-left vertex, which is not listed among the given angles. So the angles are $129^\circ$, $117^\circ$, $40^\circ$, $40^\circ$, $t$, and $137^\circ$ is given but the right angle is $90^\circ$ at the bottom-left vertex. 6. **Re-examine the problem:** The pentagon has five angles: $129^\circ$, $117^\circ$, $t$, $137^\circ$, and two $40^\circ$ angles. The right angle is indicated at the bottom-left vertex, so one of the $40^\circ$ angles must be $90^\circ$ instead. Since the problem says the small square indicates a right angle, one of the $40^\circ$ angles is actually $90^\circ$. 7. **Sum the known angles including the right angle:** Replace one $40^\circ$ with $90^\circ$: $$129^\circ + 117^\circ + 90^\circ + 40^\circ + 137^\circ + t$$ But this is six angles, which is impossible for a pentagon. So the problem must have five angles: $129^\circ$, $117^\circ$, $t$, $137^\circ$, and one $40^\circ$ angle. The other $40^\circ$ angle is likely the right angle, so it must be $90^\circ$ instead. 8. **Sum of the five angles:** The five angles are $129^\circ$, $117^\circ$, $t$, $137^\circ$, and $40^\circ$ (assuming the right angle is $40^\circ$ is incorrect, so the right angle is $90^\circ$ instead of $40^\circ$). So the angles are $129^\circ$, $117^\circ$, $t$, $137^\circ$, and $90^\circ$. 9. **Calculate $t$:** Using the sum of interior angles: $$129 + 117 + t + 137 + 90 = 540$$ Simplify: $$473 + t = 540$$ Solve for $t$: $$t = 540 - 473 = 67$$ 10. **Final answer:** The size of angle $t$ is $67^\circ$.