If in ΔABC $∠A+∠B=90°$ and $\sin B =\frac{4}{5}$, then find the value of $\cos A$.

1 $\frac{3}{5}$ $\frac{4}{5}$ 0

$\frac{4}{5}$

∠A+∠B=90°   ( given ) We know , In a triangle ABC ∠A + ∠B + ∠C = 180° 90° + ∠C = 180° ∠C = 90° So, ABC is a right angle triangle. sinB = \(\frac{4}{5}\) { Sin B = \(\frac{P}{H}\) } By using pythagoras theorem, P² + B² = H² 4² + B² = 5² 16 + B² = 25 B = 3 Now, cosA = \(\frac{B}{H}\) = \(\frac{4}{5}\)