Question

In ΔABC, m∠B = 90. If BC=3 and AC=5, find all the six trigonometric ratios of ∠A.

Answer 1

in ∆ABC, m∠B = 90°. If BC=3 and AC=5
from Pythagoras theorem,
AC² = AB² + BC²
5² = AB² + 3²
25 - 9 = AB²
16 = AB²
AB = ± 4 but AB is length so, AB ≠ -4
hence, AB = 4, BC = 3 and AC = 5

we know that ,
$\bf{sin\theta=\frac{perpendicular}{hypotenuse}}$

$\bf{cos\theta=\frac{base}{hypotenuse}}$

$\bf{tan\theta=\frac{perpendicular}{base}}$

so, sinA = BC/AC = 3/5 and cosecA = 1/sinA = 1/{3/5} = 5/3

cosA = AB/AC = 4/5 and secA = 1/cosA = 1/{4/5} = 5/4

tanA = BC/AB = 3/4 and cotA = 1/tanA = 1/{3/4} = 4/3

Answer 2

Step-by-step explanation:

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