CAT 2023 Question Paper | Quant Slot 3

CAT 2023 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 17 : 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 \(\mathrm{B}\) on \(\mathrm{AC}\). If \(\mathrm{AD}\) and \(\mathrm{BE}\) intersect at \(\mathrm{O}\) such that \(\angle \mathrm{AOB}=105^{\circ}\), then \(\frac{A D}{B E}\) equals

Video Explanation

Explanatory Answer

Area of triangle ABC = \( \frac { 1 } { 2 } \times A D \times B C = \frac { 1 } { 2 } \times B \in \times A C \)
\( \frac { A D } { B E } = \frac { A C } { B C } \)
\( \angle B O A + \angle A O E = 180 ^ { \circ } \quad ( B E \) is a straight line \( ) \)
\( \angle A O E = 75 ^ { \circ } \)
\( \angle A O E + \angle O A E + \angle A E O = 180 ^ { \circ } \) (Sum of internal angles of a triangle)
\( \angle O A E = 15 ^ { \circ } \)
Since \( \triangle B O D \cong \triangle A O E , \angle O B D = 15 ^ { \circ } \)
\( \angle B C A = \angle B C E = 180 ^ { \circ } - ( \angle E B C + \angle B E C ) = 75 ^ { \circ } \)
By symmetry, \( \angle A B D = \angle B C A = 75 ^ { \circ } \)
\( \Rightarrow \angle A B C = 75 ^ { \circ } - 15 ^ { \circ } = 60 ^ { \circ } \)
In \( \triangle A B E \), \( B E = A B \times \cos \cos \left( 60 ^ { \circ } \right) \)
In \( \triangle B E C \), \( B E = B C \times \cos \cos \left( 15 ^ { \circ } \right) \)
\( A B \times \cos \cos \left( 60 ^ { \circ } \right) = B C \cos \cos \left( 15 ^ { \circ } \right) \)
Since AB = AC, \( \frac { A C } { B C } = \frac { \cos \cos \left( 15 ^ { \circ } \right) } { \cos \cos \left( 60 ^ { \circ } \right) } = 2 \cos \cos \left( 15 ^ { \circ } \right) \)

The question is " 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 \(\mathrm{B}\) on \(\mathrm{AC}\). If \(\mathrm{AD}\) and \(\mathrm{BE}\) intersect at \(\mathrm{O}\) such that \(\angle \mathrm{AOB}=105^{\circ}\), then \(\frac{A D}{B E}\) equals "

The answer is '\(2 \cos 15^{\circ}\)'

Choice C is the correct answer.

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