0 0 votes In triangle $\mathrm{A B C}$, altitudes $\mathrm{A D}$ and $\mathrm{B E}$ are drawn to the corresponding bases. If $\angle \mathrm{B A C}=45^{\circ}$ and $\angle \mathrm{A B C}=\theta$, then $\frac{A D}{B E}$ equals $\sqrt{2} \cos \theta$ $1$ $\sqrt{2} \sin \theta$ $\frac{(\sin +\cos )}{\sqrt{2}}$ Quantitative Aptitude cat2022-set2 quantitative-aptitude + – admin 5.3k points 1.3k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.