NTSE Stage 2 SAT 2015 Question Paper

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Question : 54 of 100

Marks: +1, -0

In the figure, ABC is a triangle in which AD bisects ∠A, AC = BC, ∠B = 72° and CD = 1cm. Length of BC (in cm) is

1
$1/2$
${√5-1}/2$
${√3-1}/2$

Solution:

\m/\n=1/{\m-1}

∴

\m^2-1=\n

Now from cosine rule :

\cos 72°={\m^2+\n^2-\m^2}/{2\m\n}

∴

\cos 72°=\n/{2\m}

∴

\n=2\m\cos 72°

∴

\m^2-\m=2\m \cos 72°

(∵

\n=\m^2-\m

)
∴

m^2-2\m\cos 72°-\m=0

∴

\m^2=\m(2 \cos 72°+1)

∴

\m=2\cos 72°+1

∴

\m=2({√5-1}/4)+1

(∵

\cos 72°={√5-1}/4

)

={√5-1}/2+1={√5-1+2}/2

={1+√5}/2

Now

\BD=\m-1

={1+√5}/2-1={√5-1}/2

∴ Option (c) is correct.

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