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CAT 2023 Question Paper | Quant Slot 3

CAT Previous Year Paper | CAT Quant Questions | Question 17

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

  1. \(2 \sin 15^{\circ}\)
  2. \(\cos 15^{\circ}\)
  3. \(2 \cos 15^{\circ}\)
  4. \(\sin 15^{\circ}\)

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Explanatory Answer

Altitudes of an equilateral triangle.
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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