b) Ta có:

`hat{SIR} = 90^@ - 40^@ = 50^@`

`=> hat{SIS_1} = 50^@ * 2 = 100^@`

mà `hat{AIS_2} = 40^@`

`=> hat{IS_2 A } = 180^@ - 40^@- 90^@ = 50^@`

`=> hat{I S_2 S_1} = 90^@ - 50^@ = 40^@`

`=> hat{I S_2 S_1} =  hat{H S_2 S_1} = 40^@`

Vì `hat{H S_2 S_1} = 40^@`

`=> hat{G_1 I S_1} = hat{H S_2 S_1} = 40^@` ( đồng vị )

Vậy `I S_1`//`S_2 H`