1. **State the problem:** Simplify the expression $$m = \frac{\sqrt{6} \sec 45^\circ \sec 30^\circ + 5 (\sin 37^\circ + \sin 53^\circ)}{\tan 45^\circ + 3 \sec 53^\circ}$$. 2. **Recall trigonometric values and identities:** - $\sec \theta = \frac{1}{\cos \theta}$ - $\tan 45^\circ = 1$ - $\sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$ - $\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$ - $\sin 37^\circ + \sin 53^\circ = 1$ (since $37^\circ + 53^\circ = 90^\circ$ and $\sin 37^\circ = \cos 53^\circ$) - $\sec 53^\circ = \frac{1}{\cos 53^\circ}$ 3. **Calculate numerator:** $$\sqrt{6} \times \sec 45^\circ \times \sec 30^\circ = \sqrt{6} \times \sqrt{2} \times \frac{2}{\sqrt{3}}$$ Simplify step-by-step: $$= \sqrt{6} \times \sqrt{2} \times \frac{2}{\sqrt{3}} = \frac{2 \sqrt{6 \times 2}}{\sqrt{3}} = \frac{2 \sqrt{12}}{\sqrt{3}}$$ Since $\sqrt{12} = 2 \sqrt{3}$: $$= \frac{2 \times 2 \sqrt{3}}{\sqrt{3}} = \frac{4 \sqrt{3}}{\sqrt{3}}$$ Cancel $\sqrt{3}$: $$= 4 \cancel{\sqrt{3}} / \cancel{\sqrt{3}} = 4$$ Add the second term in numerator: $$5 (\sin 37^\circ + \sin 53^\circ) = 5 \times 1 = 5$$ Total numerator: $$4 + 5 = 9$$ 4. **Calculate denominator:** $$\tan 45^\circ + 3 \sec 53^\circ = 1 + 3 \times \frac{1}{\cos 53^\circ}$$ Approximate $\cos 53^\circ \approx 0.6018$: $$= 1 + 3 \times \frac{1}{0.6018} = 1 + 3 \times 1.662 = 1 + 4.986 = 5.986$$ 5. **Final value:** $$m = \frac{9}{5.986} \approx 1.503$$ **Answer:** $$m \approx 1.503$$