NIMCET 2019 Question Paper with solutions

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Question : 4 of 120

Marks: +1, -0

If Δ =

a^2-(b - c)^2

, where Δ is the area of the Δ ABC, then tan A equals

$15/16$
$8/15$
$8/17$
$1/2$

Solution:

Given

\Δ=a^2-(b - c)^2

⇒√{s(s-a)(s-b)(s-c)}

=(a-b+c)(a+b-c)

⇒√{s(s-a)(s-b)(s-c)}=2a-2b-2c

⇒√{s(s-a)(s-b)(s-c)}=4(s-b)(s-c)

⇒1/4={(s-b)(s-c)}/√{s(s-a)(s-b)(s-c)}

⇒{1}/{4}=√{\;\;{(s-b)(s-c)}/{s(s-a)}}=\tan\;\;{A}/{2}

⇒\tan\;\;{A}/{2}=\;\;{1}/{4}

Now,

\tanA=\;\;{2\tanA{/}2}/{1-\tan^{2}A{/}2}=\;\;{2×\;\;{1}/{4}}/{1-(\;\;{1}/{4})^{2}}=8/15

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