Solution:-

In the table of natural tangent, look for a value (≤ 1.7451) which is sufficiently close to 1.7451.

We find the value 1.7391 occurs in the horizontal line beginning with 60^o and in the column headed by 6’ and in the mean difference, we see 1.7451 – 1.7391 = .0060 in the column of 5’.

So we get, θ = 60^o 6’ + 5’ = 60^o 11’.