Question

for what value of x is sin x = cos 19°, where 0°<x<90°​

Answer 1

Answer:

Sin x = cos 19° ; 0° < x < 90°

cos is the complementary of sine. This means that the value of x in sin x and the value of 19° in cos 19° adds up to 90°

x + 19° = 90°

x = 90° - 19°

x = 71°

sin 71° = cos 19°

0.9455 = 0.9455

Value of x = 71°

Answer 2

Answer:

Your Question :--- for what value of x is sin x = cos 19°, where 0°<x<90

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we know that ,

sin@ = cos(90-@)

so,

sinx = cos(90-x)

putting this value in given question we get,

cos(90-x) = cos19°

90-x = 19°

X = 90°-19° = 71° (Ans)

Some additional complementary angles ratios we must aware of :----

(i) sin (90° - θ) = cos θ

(ii) cos (90° - θ) = sin θ

(iii) tan (90° - θ) = cot θ

(iv) cot (90° - θ) = tan θ

(v) sec (90° - θ) = csc θ

(vi) csc (90° - θ) = sec θ