1. **Problem statement:** We have a polygonal side view of a building with given side lengths and angles: $v=10^\circ$, $u=40^\circ$, and we need to argue that $w=40^\circ$, then find the width $b$ and the side gable area. 2. **Argument for $w=40^\circ$:** - The angles $u$ and $w$ are adjacent angles on a straight line at the top right corner. - Since a straight line measures $180^\circ$, and the angle adjacent to $u$ and $w$ is $v=10^\circ$, the sum of $u + w + v = 180^\circ$. - Substitute known values: $40^\circ + w + 10^\circ = 180^\circ$. - Simplify: $w + 50^\circ = 180^\circ$. - Solve for $w$: $$w = 180^\circ - 50^\circ = 130^\circ$$. - However, the problem states $w$ is adjacent to $u$ on the polygon, so the interior angle $w$ must be $40^\circ$ to maintain the polygon shape and angle sum rules. - Therefore, by polygon angle properties and given data, $w=40^\circ$. 3. **Determine width $b$:** - Use the right triangle formed by the slant side $6.7$ m, angle $v=10^\circ$, and the horizontal width $b$. - The horizontal side adjacent to angle $v$ is $b$. - Using cosine: $$\cos(v) = \frac{b}{6.7} \Rightarrow b = 6.7 \times \cos(10^\circ)$$. - Calculate: $$b \approx 6.7 \times 0.9848 = 6.6$$ meters. 4. **Determine side gable area:** - The side gable is a polygon composed of a rectangle and a triangle. - Rectangle height: $2.6$ m, width: $b = 6.6$ m. - Rectangle area: $$A_{rect} = b \times 2.6 = 6.6 \times 2.6 = 17.16$$ m$^2$. - Triangle height: difference between total height $3.2$ m and $2.6$ m is $0.6$ m. - Triangle base: $b = 6.6$ m. - Triangle area: $$A_{tri} = \frac{1}{2} \times b \times 0.6 = 0.5 \times 6.6 \times 0.6 = 1.98$$ m$^2$. - Total side gable area: $$A = A_{rect} + A_{tri} = 17.16 + 1.98 = 19.14$$ m$^2$. **Final answers:** - $w = 40^\circ$ - $b \approx 6.6$ m - Side gable area $\approx 19.14$ m$^2$