Observe and then find `/_P`.

Explanation

We have,
triangleABC and trianglePQR,
`/_A = 80°`,
`/_B = 60°`,
AB =` 3.8` units, BC = `6 `units, AC = `3sqrt3` units
PR =` 6sqrt3` units, PQ = `12 `units, RQ = `7.6` units.
Now,
In `triangleABC`,
`/_A + /_B + /_C = 180^@` [Angle Sum Property]
`80^@ + 60^@ + /_C = 180^@`
`140^@ + /_C = 180^@`
`/_C = 180^@ - 140^@`
∴ `/_C = 40^@`
Now,
If we can prove that `triangleABC `and `trianglePQR` are similar then we can say that their angles will be equal.
Now,
`(AB)/(RQ) = (3.8/7.6) = (1/2)`
`(BC)/(PQ) = (6/12) = (1/2)`
`(AC)/(PR) = (3sqrt3)/(6sqrt3) = (1/2)`
Hence,
`(AB)/(RQ) = (BC)/(PQ) = (AC)/(PR)`
[When the ratio of sides of two triangles are equal, they are similar.]
∴ `triangleABC ~ triangleRQP` (By S.S.S Similarity)
Then,
`/_C = /_P` [Corresponding Parts of Similar Triangles]
We know that,
`/_C = 40^@`
∴ `/_P = 40^@`
Hence,
`/_P = 40^@`