Question

Question 1.
Given: In AABC, 2B=90°.
If 15 cot A=8, Find sin A
and sec A.​

Answer 1

Given,

15cotA=8

cotA= 15/8

=> tanA= 8/15

-------(tanA= cotA=1)

We know that,

tanθ= adjacentSide,oppositeSide

Consider the attached figure, triangleABC

From Pythagoras theorem,

2AC =2AB +2BC

2AC =8(2)+15(2)

2AC=64+225=289

AC=17

cosA= Hypotenuse

adjacentSide = AC/AB = 17/8

secA= cosA 1

= 17/8= 1

8/17

sinA= Hypotenuse

oppositeSide = AC/BC

= 17/15

solution

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-

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

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

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

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

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

_____________

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

$\rm { \qquad • Hypotenuse = 17 }$

$\qquad$

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

$\qquad$

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