In the given triangle $AB=3cm$ and $AC=5cm$. Find all trignometric ratios.
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$\sin {\theta}=\cfrac{P}{H}=\cfrac{4}{5}$
$\cos {\theta}=\cfrac{B}{H}=\cfrac{4}{5}$
$\tan {\theta}=\cfrac{P}{B}=\cfrac{4}{3}$
$\cot{\theta}=\cfrac{B}{P}=\cfrac{3}{4}$
$\sec {\theta}=\cfrac{H}{B}=\cfrac{5}{3}$
$co\sec {\theta}=\cfrac{H}{P}=\cfrac{5}{4}$ -
$\sin {\theta}=\cfrac{P}{H}=\cfrac{4}{5}$
$\cos {\theta}=\cfrac{B}{H}=\cfrac{2}{5}$
$\tan {\theta}=\cfrac{P}{B}=\cfrac{4}{3}$
$\cot{\theta}=\cfrac{B}{P}=\cfrac{3}{4}$
$\sec {\theta}=\cfrac{H}{B}=\cfrac{5}{3}$
$co\sec {\theta}=\cfrac{H}{P}=\cfrac{5}{4}$ -
$\sin {\theta}=\cfrac{P}{H}=\cfrac{4}{5}$
$\cos {\theta}=\cfrac{B}{H}=\cfrac{1}{5}$
$\tan {\theta}=\cfrac{P}{B}=\cfrac{4}{3}$
$\cot{\theta}=\cfrac{B}{P}=\cfrac{3}{4}$
$\sec {\theta}=\cfrac{H}{B}=\cfrac{5}{3}$
$co\sec {\theta}=\cfrac{H}{P}=\cfrac{5}{4}$ -
$\sin {\theta}=\cfrac{P}{H}=\cfrac{4}{5}$
$\cos {\theta}=\cfrac{B}{H}=\cfrac{3}{5}$
$\tan {\theta}=\cfrac{P}{B}=\cfrac{4}{3}$
$\cot{\theta}=\cfrac{B}{P}=\cfrac{3}{4}$
$\sec {\theta}=\cfrac{H}{B}=\cfrac{5}{3}$
$co\sec {\theta}=\cfrac{H}{P}=\cfrac{5}{4}$
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AI explanation
In the right-angled triangle, the hypotenuse AC is 5 cm and the base AB is 3 cm. Using the Pythagorean theorem, the perpendicular side BC is calculated as the square root of (5^2 - 3^2), which is 4 cm. Based on these sides, the sine of the angle theta is Perpendicular/Hypotenuse = 4/5, and the cosine of theta is Base/Hypotenuse = 3/5. The other trigonometric ratios follow as tan(theta) = 4/3, cot(theta) = 3/4, sec(theta) = 5/3, and cosec(theta) = 5/4, which matches the required option.