Question

Given 15 cot A=8 find sin A and sec A

Answer 1

cot A = 8 / 5 ( cot = b/ p )

Now;

p^2+b^2 = h^2


(5)^2 + (8)^2= h^2


25 + 64 = h^2

h^2 = 89

h = _/89

sin A = 5 / _/ 89( sinA = p/ h )

secA = _/89 / 8 ( h/ b)

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Answer 2

Given that,

$\rm { \qquad • 15 cot \: A = 8 }$

_________

We have to find,

$\rm { \qquad • sin \: A \: and \: sec \: A }$

_________

Solution,

If $\rm {15 cot \: A = 8 }$, then $\rm { cot \: A = \dfrac{8}{15} }$

We know that ,

$\rm { \qquad • cot \: A = \dfrac{Base}{Perpendicular}}$

Hence,

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

____________

Applying pythagoras property-

$\rm\red { {H}^{2} = {B}^{2}+{P}^{2} }$

$\rm\leadsto { {H}^{2} = {8}^{2}+{15}^{2} }$

$\rm\leadsto { {H}^{2} = 64+ 225}$

$\rm\leadsto { {H }^{2} = 289}$

$\leadsto\rm\blue { H = \sqrt{289} = 17}$

_____________

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

$\rm { \qquad • Hypotenuse = 17 }$

$\qquad$

$\rm\red { \leadsto sin \: A = \dfrac{ Perpendicular}{Hypotenuse} =\dfrac{15}{17} }$

$\qquad$

$\rm\red { \leadsto sec \: A = \dfrac{ Hypotenuse}{Base} =\dfrac{17}{8} }$