triangle ABC

<B =?

ab =root 7

ac =4

bc =6

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Given triangle ABC with AB=`sqrt(7)` ,AC=4 and BC=6 find the measure of angle B:

Use the law of cosines: `c^2=a^2+b^2-2abcosC`

`(sqrt(7))^2=4^2+6^2-2(4)(6)cosB`

7=16+36-48cosB

-45=-48cosB

`cosB=15/16`

`B=cos^(-1)(15/16)~~20.4^@`

**The angle has measure of approximately 20.4 degrees.**

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