Question

How do you find the measure of angle B for the triangle with the following dimensions. A = 71° , a = 9.3, b = 8.5?

Answer 1

`"B=59.8"^"o"`
Use the law of sines: `"a"/("sinA")="b"/"sinB"="c"/"sinC"`.

Explanation:

Law of sines: `"a"/("sinA")="b"/"sinB"="c"/"sinC"`.

`"A=71"^"o"`
`"a"=9.3`
`"b"=8.5`
`"B=?"`

Since there are no values for side c and angle C, we can ignore this ratio. Since we have the sides a and b, but don't have angle B, we can invert the ratios for the law of sines.

`("sinA")/"a"=("sinB")/"b"` =

`("sin71"^"o")/8.5=("sinB")/9.3`

Divide `"sinA"` by side `"a"`.

`0.101669=("sinB")/8.5`

Multiply both sides by `8.5`.

`0.101669xx8.5="sinB"`

`0.8641865="sinB"`

Switch sides.

`"sinB"=0.8641865`

Determine angle `"B"` by taking the inverse sin of `0.85`.

`"B=sin"^(-1)(0.8641865)`

`"B=59.8"^"o"`