In triangle ABC, angle C is 90⁰, sine A = 1/3. Find the sine and cosine of the outer corner at vertex A.

The sum of the inside angle A and outside angle F is 180 °.

<A + <F = 180 °;

<F = 180 – <A;

sin (<F) = sin (180 – <A) = sin (180 – <A) = sin (<A) = 1/3.

A reduction formula is applied to the expression sin (180 – <A).

cos (<F) = √ (1 – sin ^ 2 (<A)) = √ (1- (1/3) ^ 2) = √ (1 – 1/9) = √8 / 9 = 2√2 / 3;

Answer: sin (<F) = 1/3, cos (F) = 2√2 / 3.