WAEC 2022 · Paper 1 · Q29

Trigonometric ratios

Given that $\sin(5x - 28)^\circ = \cos(3x - 50)^\circ$ , $0^\circ \le x \le 90^\circ$ , find the value of $x$ .

Worked solution (try it first)

$\sin A = \cos B$ when $A$ and $B$ are complementary, $A + B = 90^\circ$ .
So $(5x - 28) + (3x - 50) = 90$ .
Collect terms: $8x - 78 = 90$ , so $8x = 168$ .
Divide by 8: $x = 21$ , option C.
(Check: the angles are $77^\circ$ and $13^\circ$ , which add to $90^\circ$ .)

Report a problem with this question