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CAT 2023 Slot 3 — QA Question 17

Geometric CentersEasy

Let triangle ABC be isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that ∠AOB = 105°, then AD/BE equals?

Answer & solution

Correct answer: 2 cos 15°

  • 2 cos 15°

  • B

    cos 15°

  • C

    2 sin 15°

  • D

    sin 15°

Solution

Easy

Both altitudes give the same area of △ABC\triangle ABC, so ADBE=ACBC\dfrac{AD}{BE}=\dfrac{AC}{BC}. Find the angles from ∠AOB=105∘\angle AOB=105^\circ, then turn ACBC\dfrac{AC}{BC} into a sine ratio using the area formula 12 absin⁡C\tfrac12\,ab\sin C.

1

Find the angles. BE⊥ACBE\perp AC and AD⊥BCAD\perp BC. In △OBD\triangle OBD, ∠ODB=90∘\angle ODB=90^\circ and ∠BOD=180∘−105∘=75∘\angle BOD=180^\circ-105^\circ=75^\circ:

∠OBD=90∘−75∘=15∘∠ECB=90∘−15∘=75∘=∠C∠B=∠C=75∘,∠A=180∘−75∘−75∘=30∘\begin{aligned} &\angle OBD=90^\circ-75^\circ=15^\circ\\ &\angle ECB=90^\circ-15^\circ=75^\circ=\angle C\\ &\angle B=\angle C=75^\circ,\quad \angle A=180^\circ-75^\circ-75^\circ=30^\circ \end{aligned}
2

Equate the two area expressions using the two altitudes:

[△ABC]=12⋅AD⋅BC=12⋅BE⋅AC⇒ ADBE=ACBC\begin{aligned} &[\triangle ABC]=\tfrac12\cdot AD\cdot BC=\tfrac12\cdot BE\cdot AC\\ &\Rightarrow\ \frac{AD}{BE}=\frac{AC}{BC} \end{aligned}
3

Use the side–angle area formula (AB=ACAB=AC, equal sides) to get ACBC\dfrac{AC}{BC}:

[△ABC]=12⋅AB⋅ACsin⁡A=12⋅AB⋅BCsin⁡B⇒ ACBC=sin⁡Bsin⁡A=sin⁡75∘sin⁡30∘=sin⁡75∘12=2sin⁡75∘\begin{aligned} &[\triangle ABC]=\tfrac12\cdot AB\cdot AC\sin A=\tfrac12\cdot AB\cdot BC\sin B\\ &\Rightarrow\ \frac{AC}{BC}=\frac{\sin B}{\sin A}=\frac{\sin 75^\circ}{\sin 30^\circ}=\frac{\sin 75^\circ}{\tfrac12}=2\sin 75^\circ \end{aligned}
4

Convert to cosine since sin⁡75∘=cos⁡15∘\sin 75^\circ=\cos 15^\circ:

ADBE=ACBC=2sin⁡75∘=2cos⁡15∘\frac{AD}{BE}=\frac{AC}{BC}=2\sin 75^\circ=2\cos 15^\circ
ADBE=2cos⁡15∘\frac{AD}{BE}=\mathbf{2\cos 15^\circ}

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CAT 2023 Slot 3 QA Q17 — Geometric Centers solution | TheCATExam