Question

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

Answer:

kindly see the answer below...

Step-by-step explanation:

HERE,

given that,

$15 \cot( \alpha ) = 8$

$or \: \: \cot( \alpha ) = \frac{8}{15}$

then, according to the triangle rule,

$perpendicular = 15$

$base = 8$

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

$= \sqrt{64 + 225}$

$= \sqrt{289}$

$= 17$

Therefore,

$\sin( \alpha ) = \frac{perpendicular}{hypotenuse}$

$= \frac{15}{17}$

and,

$\sec( \alpha ) = \frac{hypotenuse}{base}$

$= \frac{17}{8}$

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Answer 2

cotA = 8/15

cotA= 1/ tan. tanA= 1/cot

so, tanA= 1/(8/15).

= 15/8

we know tanA= sin/ cos

so 15/8= sin/ cos

so, sinA= 15

if we take triangle ABC

A is the angle

so,AB= 8 (given)

BC= 15( we find)

then we find CA

• we know phytagaros Theron= AB2+ BC2= AC2

8 square+ 15 square= AC2

64+ 225=AC2

289square= AC2

17= AC

then AC = 17

sinA= opposite side of A/ hypotenuse

15/17

secA= hypotenuse/ adjacent to A

8/17