1 1 vote Let $\triangle A B C$ be an isosceles triangle such that $A B$ and $A C$ are of equal length. $A D$ is the altitude from $A$ on $B C$ and $B E$ is the altitude from $B$ on $A C$. If $A D$ and $B E$ intersect at $O$ such that $\angle A O B=105^{\circ}$, then $\frac{A D}{B E}$ equals$\sin 15^{\circ}$$\cos 15^{\circ}$$2 \cos 15^{\circ}$$2 \sin 15^{\circ}$ Quantitative Aptitude cat2023-set3 quantitative-aptitude geometry + – admin 5.3k points 1.2k views answer comment Share Follow Print See 1 comment 1 1 comment reply rhl 642 points commented Aug 10, 2025 reply Follow flag Make a diagram and use exterior angle property, opposite angle theorem to find out some missing angles. after wards in the $2$ Right angled triangles i used $base=hypotenuse*cos\theta$ and $height=hpyotenuse*sin\theta$ , where $\theta$ is the anlge between hypotenuse and base. Finally i got $\frac{sin75^{\circ}}{sin30^{\circ}}$ I had to take help from chatgpt after this basically there is an identity that $sin\theta=cos(90-\theta)$ and $sin30^{\circ}=\frac{1}{2}$ Hence answer is $2*cos15^{\circ}$ 0 0 replyShare Please log in or register to add a comment.