In - ABC- - A - 120- BC-CA-20 - AB-BC-21-then

The correct option is A AB gt; AC From the fig, a+b=20 (i) c+a=21 (ii) c b=1 c=b+1 i.e,c gt;b AB gt;AC ...

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Standard XII

Mathematics

Existence of Limit

In Δ ABC, ∠...

Question

$I n Δ A B C$ , $∠ A = 120$ , $B C + C A = 20$ , $A B + B C = 21$ ,then

A B

A C

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△ A B C

is isosceles

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A r e a o f Δ A B C = 14 \sqrt{3}

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Solution

The correct option is A $A B$ > $A C$
$F r o m t h e f i g,$
$a + b = 20 - (i)$
$c + a = 21 - (i i)$
$⟹ c - b = 1$
$⟹ c = b + 1 i . e, c > b$
$⟹ A B > A C$
$△ = \frac{1}{2} b c sin A = \frac{1}{2} b (b + 1) sin 120$
$= \frac{1}{2} b (b + 1) \frac{\sqrt{3}}{2}$
$∴ △ = b (b + 1) \frac{\sqrt{3}}{4}$
$⟹ A B > A C$

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InΔABC, ∠A=120, BC+CA=20 , AB+BC=21,then

The correct option is A AB>ACFromthefig,a+b=20−(i)c+a=21−(ii)⟹c−b=1⟹c=b+1i.e,c>b⟹AB>AC△=12bcsinA=12b(b+1)sin120=12b(b+1)√32∴△=b(b+1)√34⟹AB>AC

$I n Δ A B C$ , $∠ A = 120$ , $B C + C A = 20$ , $A B + B C = 21$ ,then