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In /\ABC,angleB=42^(@),angleC=30^(@) , a...

In `/_\ABC,angleB=42^(@),angleC=30^(@)` , and AB=100. The length of BC is

A

47.6

B

66.9

C

133.8

D

190.2

Text Solution

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The correct Answer is:
D
`angleA=108^(@)`. Law of sines: `(BC)/(sin108^(@))=(100)/(sin30^(@))`. Therefore, BC`=(100sin108^(@))/(sin30^(@))=190.2`. [6]
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