Question

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

Answer 1

Answer:D

15cot A =8

cotA =8/15

15=perpendicular

8=base

Now by Pythagoras theorem

(H)²=(P)²+(B)²

(H)²= 225+64

Therefore hypotenuse is equal to 17

Now ,

sinA= 15/17

sec A=17/8.

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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-

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

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

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

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

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

_____________

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

$\rm { \qquad • Hypotenuse = 17 }$

$\qquad$

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

$\qquad$

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