Question

How do you solve the triangle given ∠C = 68°, ∠B = 24°, b = 11?

Answer 1

`/_A=88°, a= 27,c=25`

Explanation:

First draw a triangle then find `/_A`. The diagram will enable you to know exactly what you are ealing with.

`/_A+/_B+/_C=180°`

`/_A=180°-/_B-/_C`

`/_A= 180°-24°-68°`

`/_A=88°`

Now that we have `/_A`, we can use the law of sine to find a which is the length opposite `/_A`.

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

`a*sinB=b*sinA`

`(a*cancel(sinB))/cancelsinB=(b*sinA)/sinB`

`a=(11*sin88°)/(sin24°)`

`a=27`

Now use the sine rule again to colve for `c`

`(sin C)/c=(sinB)/b`

`c=(b*sinC)/sinB`

`c=(11*sin68°)/sin24°`

`c=25`