JAMB - Mathematics (2019 - No. 8)
Given \(\sin 58° = \cos p°\), find p.
48°
58°
32°
52°
Тайлбар
\(\sin \theta = \cos (90 - \theta)\)
\(\sin \theta = \cos (90 - 58)\)
= \(\cos 32\)
\(\sin \theta = \cos (90 - \theta)\)
\(\sin \theta = \cos (90 - 58)\)
= \(\cos 32\)