Fresh day gurlll_zelight!, let's solve. Step 1: Calculate the angle $\beta$ in triangle 'b'. In the right-angled triangle 'b', the side opposite to angle $\beta$ is 9, and the hypotenuse is 20. We use the sine function. $$ \sin(\beta) = \frac{\text{opposite}}{\text{hypotenuse}} $$ $$ \sin(\beta) = \frac{9}{20} $$ $$ \sin(\beta) = 0.45 $$ To find $\beta$, we take the inverse sine: $$ \beta = \arcsin(0.45) $$ $$ \beta \approx 26.7436^\circ $$ Rounding to the nearest 0.1 degree: $$ \beta \approx \boxed{26.7^\circ} $$ Step 2: Calculate the angle $\delta$ in triangle 'd'. In the right-angled triangle 'd', the side opposite to angle $\delta$ is 8, and the side adjacent to angle $\delta$ is 6. We use the tangent function. $$ \tan(\delta) = \frac{\text{opposite}}{\text{adjacent}} $$ $$ \tan(\delta) = \frac{8}{6} $$ $$ \tan(\delta) = \frac{4}{3} $$ To find $\delta$, we take the inverse tangent: $$ \delta = \arctan\left(\frac{4}{3}\right) $$ $$ \delta \approx 53.1301^\circ $$ Rounding to the nearest 0.1 degree: $$ \delta \approx \boxed{53.1^\circ} $$ Send me the next one 📸