Question

In a right angle triangle, if angle A is acute and cot A = 4/3 , find sin A, cos A and tan A

Answer 1

Answer:

sin a = 3/5

cos a = 4/5

tan a = 3/4

hope its helpful

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Answer 2

Solution :

In a right angle triangle, if angle A is acute angle & cot A = 4/3,so according to the question here attach a diagram of right angle;

$\underline{\bf{Explanation\::}}}}$

As we know that;

$\boxed{\bf{cot\theta=\frac{Base}{Perpendicular} }}}$

$\longrightarrow\sf{cot \:A= \dfrac{4}{3} =\dfrac{AC}{BC} }$

$\underline{\boldsymbol{By\:using\:pythagoras\:theorem\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }\\\\\mapsto\sf{(AB)^{2} = (AC)^{2} + (BC)^{2} } \\\\\mapsto\sf{(AB)^{2} = (4)^{2} + (3)^{2} }\\\\\mapsto\sf{(AB)^{2} = 16 + 9}\\\\\mapsto\sf{(AB)^{2} = 25}\\\\\mapsto\sf{AB = \sqrt{25} }\\\\\mapsto\bf{AB = 5\:unit}$

Thus;

$\bullet\:\sf{sin\:A=\dfrac{Perpendicular}{Hypotenuse} =\dfrac{BC}{AB} = \boxed{\bf{\frac{3}{5} }}}\\\\\\\bullet\sf{cos\:A=\dfrac{Base}{Hypotenuse} =\dfrac{AC}{AB} = \boxed{\bf{\frac{4}{5} }}}\\\\\\\bullet\sf{tan\:A=\dfrac{Perpendicular}{Base} =\dfrac{BC}{AC} = \boxed{\bf{\frac{3}{4} }}}$