Let \triangle A B C be an 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 \angle A O B=105^{\circ}, then \frac{A D}{B E} equals

CAT 2023 Slot 3 Solved Paper

Question : 61 of 66

Marks: +1, -0

Let

△ A B C

be an isosceles triangle such that

{\AB}

and

{\AC}

are of equal length.

{\AD}

is the altitude from

{\A}

{\BC}

and

{\BE}

is the altitude from

{\B}

{\AC}

. If

{\AD}

and

{\BE}

intersect at

{\O}

such that

∠ A O B=105^{∘}

, then

\;{A D}/{B E}

equals

Solution:

Given that

A B=A C⇒

Angle

C=

Angle

B

(1)

A D

and

B E

are altitudes ⇒ they make 90 degrees with the sides
Angle

{\AOB}=105⇒

Angle EOD

=105

(Vertically Opposite Angles)
In quadrilateral DOEC
Angle

C=360-105-90-90=75⇒

Angle

B=75

(from 1)
We know that from the area of the triangle

A D * B C=B E * A C

\;⇒\;{AD}/{BE}=\;{AC}/{BC}=\;{2 R \sin \;B}/{2R \sin \;A}=\;{ \sin \;(75)}/{ \sin \;(30)}=2 \sin \;(75)=2 \cos (15)

\;{[ \sin \;( {\x})=\cos (90- {\x})]}

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