Solution:-

From the question it is given that, sin x^o = 0.67.

In the table of natural sines, look for a value (≤ 0.67) which is sufficiently close to 0.67.

We find the value 0.6691 occurs in the horizontal line beginning with 42^o and in the mean difference, we see 0.6700 – 0.6691 = .0009 in the column of 4’.

So we get, θ = 42^o + 4’ = 42^o 4’.

Then,

(i) cos x^o = cos 42^o 4′

From the table

= .7431 – .0008

= 0.7423

(ii) cos x^o + tan x° = cos 42° 4′ + tan 42° 4′

= 0.7423 + .9025

= 1.6448