Here are the solutions to the equations: a) Solve $\cos \alpha = \sin 40^\circ$ Step 1: Use the complementary angle identity $\sin x = \cos(90^\circ - x)$. $$ \sin 40^\circ = \cos(90^\circ - 40^\circ) = \cos 50^\circ $$ Step 2: Substitute this back into the equation. $$ \cos \alpha = \cos 50^\circ $$ Step 3: Solve for $\alpha$. $$ \alpha = 50^\circ $$ The solution is $\boxed{\alpha = 50^\circ}$. b) Solve $\sin \beta = \cos 80^\circ$ Step 1: Use the complementary angle identity $\cos x = \sin(90^\circ - x)$. $$ \cos 80^\circ = \sin(90^\circ - 80^\circ) = \sin 10^\circ $$ Step 2: Substitute this back into the equation. $$ \sin \beta = \sin 10^\circ $$ Step 3: Solve for $\beta$. $$ \beta = 10^\circ $$ The solution is $\boxed{\beta = 10^\circ}$. c) Solve $\sin 2\theta = \cos 3\theta$ Step 1: Use the complementary angle identity $\cos x = \sin(90^\circ - x)$. $$ \cos 3\theta = \sin(90^\circ - 3\theta) $$ Step 2: Substitute this back into the equation. $$ \sin 2\theta = \sin(90^\circ - 3\theta) $$ Step 3: Equate the angles. $$ 2\theta = 90^\circ - 3\theta $$ Step 4: Solve for $\theta$. $$ 2\theta + 3\theta = 90^\circ $$ $$ 5\theta = 90^\circ $$ $$ \theta = \frac{90^\circ}{5} $$ $$ \theta = 18^\circ $$ The solution is $\boxed{\theta = 18^\circ}$. d) Solve $\sin \gamma = \cos(\gamma + 20^\circ)$ Step 1: Use the complementary angle identity $\cos x = \sin(90^\circ - x)$. $$ \cos(\gamma + 20^\circ) = \sin(90^\circ - (\gamma + 20^\circ)) $$ $$ \cos(\gamma + 20^\circ) = \sin(90^\circ - \gamma - 20^\circ) $$ $$ \cos(\gamma + 20^\circ) = \sin(70^\circ - \gamma) $$ Step 2: Substitute this back into the equation. $$ \sin \gamma = \sin(70^\circ - \gamma) $$ Step 3: Equate the angles. $$ \gamma = 70^\circ - \gamma $$ Step 4: Solve for $\gamma$. $$ \gamma + \gamma = 70^\circ $$ $$ 2\gamma = 70^\circ $$ $$ \gamma = \frac{70^\circ}{2} $$ $$ \gamma = 35^\circ $$ The solution is $\boxed{\gamma = 35^\circ}$. e) Solve $\sin \phi - \cos \phi = 0$ Step 1: Rearrange the equation. $$ \sin \phi = \cos \phi $$ Step 2: Divide both sides by $\cos \phi$ (assuming $\cos \phi \neq 0$). $$ \frac{\sin \phi}{\cos \phi} = 1 $$ Step 3: Use the identity $\tan \phi = \frac{\sin \phi}{\cos \phi}$. $$ \tan \phi = 1 $$ Step 4: Solve for $\phi$. $$ \phi = \arctan(1) $$ $$ \phi = 45^\circ $$ The solution is $\boxed{\phi = 45^\circ}$. f) Solve $\cos \delta = \sin(\delta - 10^\circ)$ Step 1: Use the complementary angle identity $\cos x = \sin(90^\circ - x)$. $$ \cos \delta = \sin(90^\circ - \delta) $$ Step 2: Substitute this back into the equation. $$ \sin(90^\circ - \delta) = \sin(\delta - 10^\circ) $$ Step 3: Equate the angles. $$ 90^\circ - \delta = \delta - 10^\circ $$ Step 4: Solve for $\delta$. $$ 90^\circ + 10^\circ = \delta + \delta $$ $$ 100^\circ = 2\delta $$ $$ \delta = \frac{100^\circ}{2} $$ $$ \delta = 50^\circ $$ The solution is $\boxed{\delta = 50^\circ}$. Send me the next one 📸