Question

4. Given 15 cot A = 8, find sin A and sec A.​

Answer 1

Answer:

sin A =p/h=15x/17x=15/17

sec A.=h/b=17x/8x=17/8

Step-by-step explanation:

 15 cot A = 8

cot A =8/15=b/p

base=8x     perpendicular=15x

hypotenuse=$\sqrt{h^{2}+b^{2} }$

hypotenuse=$\sqrt{(15x)^{2}+(8x)^{2} }$

hypotenuse=17x

sin A =p/h=15x/17x=15/17

sec A.=h/b=17x/8x=17/8

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-

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

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

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

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

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

_____________

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

$\rm { \qquad • Hypotenuse = 17 }$

$\qquad$

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

$\qquad$

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