Question

in a triangle abc right angled at b if AB is 12 and BC is 5 find all the six trignometric ratios of angle. a​

Answer 1

Step-by-step explanation:

$in \: the \: triangle \: b \: is \: right \: angle \: so \: for \: the \: angle \: a \: \\ ab = base(b) = 12 \\ bc = perpendicular(p) = 5 \: and \: the \: \\ ac = heipotenous(h) \\ so \: h = \sqrt{ {p}^{2} + {b}^{2} } = \sqrt{ {12}^{2} + {5}^{2} } \\ = \sqrt{144 + 25} = \sqrt{169} = 13 \\ so \: all \: the \: trigonometric \: eqs \: are \\ \sin( \alpha ) = \frac{p}{h} = \frac{5}{13} \\ \cos( \alpha ) = \frac{b}{h} = \frac{12}{13} \\ \tan( \alpha ) = \frac{p}{b} = \frac{5}{12} \\ \cot( \alpha ) = \frac{b}{p} = \frac{12}{5} \\ \sec( \alpha ) = \frac{h}{b} = \frac{13}{12} \\ \csc( \alpha ) = \frac{h}{p} = \frac{13}{5}$

Answer 2

$in \: the \: triangle \: b \: is \: right \: angle \: so \: for \: the \: angle \: a \: \\ ab = base(b) = 12 \\ bc = perpendicular(p) = 5 \: and \: the \: \\ ac = heipotenous(h) \\ so \: h = \sqrt{ {p}^{2} + {b}^{2} } = \sqrt{ {12}^{2} + {5}^{2} } \\ = \sqrt{144 + 25} = \sqrt{169} = 13 \\ so \: all \: the \: trigonometric \: eqs \: are \\ \sin( \alpha ) = \frac{p}{h} = \frac{5}{13} \\ \cos( \alpha ) = \frac{b}{h} = \frac{12}{13} \\ \tan( \alpha ) = \frac{p}{b} = \frac{5}{12} \\ \cot( \alpha ) = \frac{b}{p} = \frac{12}{5} \\ \sec( \alpha ) = \frac{h}{b} = \frac{13}{12} \\ \csc( \alpha ) = \frac{h}{p} = \frac{13}{5}$