`a)`

Xét `DeltaABH` vuông tại `H` có:

`sinB=(AH)/(AB)`

`=>AH=AB.sinB`

`=>AH=10.sin60^@=10.sqrt(3)/2`

`=>AH=5sqrt(3)\ (cm)`

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`cosB=(BH)/(AB)`

`=>BH=AB.cosB`

`=>BH=10.cos60^@`

`=>BH=10.1/2=5\ (cm)`

`b)`

Xét `DeltaACH` vuông tại `H` có:

`sinC=(AH)/(AC)`

`=>AC=(AH)/sinC`

`=>AC=(5sqrt(3))/sin45^@`

`=>AC=(5sqrt(3))/(sqrt(2)/2)`

`=>AC=5sqrt(6)\ (cm)`

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`tan(C)=(AH)/(CH)`

`=>CH=(AH)/tanC`

`=>CH=(5sqrt(3))/tan45^@`

`=>CH=5sqrt(3)\ (cm)`

Ta có: `HinBC` nên:

`BC=BH+CH=5+5sqrt(3)\ (cm)`