TheCATExam
Sign in

CAT 2022 Slot 2 — QA Question 18

Basics of TrianglesEasy

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC = 45° and ∠ABC = θ, then AD/BE equals

Answer & solution

Correct answer: √2 sinθ

  • A

    1

  • B

    √2 cosθ

  • √2 sinθ

  • D

    (sinθ + cosθ)/√2

Solution

Easy

The area of △ABC\triangle ABC can be written using two different pairs of sides with their included angles (45∘45^\circ at AA, θ\theta at BB), and also via the two altitudes. Equate to get AD/BEAD/BE.

1

Area via included angles at AA and at BB:

[ABC]=12 AB⋅ACsin⁡45∘=AB⋅AC22[ABC]=12 BA⋅BCsin⁡θ\begin{aligned} &[ABC]=\tfrac12\,AB\cdot AC\sin 45^\circ=\dfrac{AB\cdot AC}{2\sqrt2}\\ &[ABC]=\tfrac12\,BA\cdot BC\sin\theta \end{aligned}
2

Equate (the common ABAB cancels):

AC22=BCsin⁡θ2 ⇒ ACBC=2 sin⁡θ\dfrac{AC}{2\sqrt2}=\dfrac{BC\sin\theta}{2}\ \Rightarrow\ \dfrac{AC}{BC}=\sqrt2\,\sin\theta
3

Area via altitudes. ADAD is the altitude to base BCBC, BEBE the altitude to base ACAC:

12 AD⋅BC=12 BE⋅AC ⇒ ADBE=ACBC\tfrac12\,AD\cdot BC=\tfrac12\,BE\cdot AC\ \Rightarrow\ \dfrac{AD}{BE}=\dfrac{AC}{BC}
4

Combine steps 2 and 3:

ADBE=2 sin⁡θ\dfrac{AD}{BE}=\sqrt2\,\sin\theta
ADBE=2 sin⁡θ\dfrac{AD}{BE}=\mathbf{\sqrt2\,\sin\theta}

Related Basics of Triangles questions

See all Triangles questions →
CAT 2022 Slot 2 QA Q18 — Basics of Triangles solution | TheCATExam