In the interval 0 x 360 find the values of x for which cos x =0.7252

Step-by-step explanation:

We have a positive value for the cosine of x, so we know that the value of x should be in the first quadrant (0 ≤ x ≤ 90) or in the fourth quadrant (270 ≤ x ≤ 360).

Now, let's find the value of x that gives cos(x) = 0.7252 using the inverse function of the cosine, that is, the arc cosine function.

The value of x can be calculated using:

Using this function in a calculator (you may find it as: ), we have that:

So this is the value of x in the first quadrant. To find the other value of x, in the fourth quadrant, that gives the same result, we just need to calculate 360° minus the value we found:

So the values of x are:

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

what "following statement"?

step-by-step explanation: