Question

How would you find the sides of a triangle given angle A=50, angle B=30, and side a=1?

Answer 1

`a = 1` (given)
`b = 0.65`
`c = 1.29`

Explanation:

To solve this, we use the Law of Sines

`b/sinB = a/sinA`

`b/(sin30^@) = 1/sin50^@`

`b = sin30^@/sin50^@`

`b ~~ 0.65`

`-------------------`

We can find angle C by doing `180^@ - 50^@ - 30^@ = 100^@`

Now let's find side c:
`c/sinC = a/sinA`

`c/(sin100^@) = 1/sin50^@`

`c = sin100^@/sin50^@`

`c ~~ 1.29`

Therefore, the lengths of the sides of the triangle are:
`a = 1` (given)
`b = 0.65`
`c = 1.29`

(`b` and `c` are rounded to the nearest hundredth)

Hope this helps!