Question

Given 15cot A = 8 find sin A sec A ​

Answer 1

$\huge\boxed{\fcolorbox{black}{red}{answer}}$

Given:-

$15 \cot(A) = 8$

To find :-

$\sin(A) \: and \: \sec(A)$

Therefore :-

$\cot(A) = \frac{8}{15}$

$= \frac{b}{p}$

Base = 8 and

Perpendicular = 15

Find hypotenuse :-

${a}^{2} = {b}^{2} + {c}^{2}$

$✏ {a}^{2} = {8}^{2} + {15}^{2} </p><p>$

$✏ {a}^{2} = 64 + 15</p><p>$

$✏a = \sqrt{64 + 225} </p><p>$

$✏a = 17</p><p>$

$\sin(A) = \frac{15}{17}$

$\sec(A) = \frac{17}{8}$

Important Formul :-

$\sin(A) = \frac{p}{h}$

$\cos(A) = \frac{b}{h}$

$\tan(A) = \frac{p}{b}$

$\cot(A) = \frac{b}{p}$

$\sec(A) = \frac{h}{b}$

$\csc(A) = \frac{h}{p}$

Answer 2

Given:

$\huge{\sf{15\: cot(A)=8}}$

To find :-

$\huge{\sf{sin(A)  and sec(A) }}$

Therefore :-

$\huge{\sf{cot(A)  = \frac{8}{15}}$

$\huge{\sf{= \frac{b}{p}}$

• Base = 8 and
• Perpendicular = 15

Find hypotenuse :-

$\huge{\sf{a^2 = b^2+c^2}}$

$\huge{\sf{a^2}={ 8^2+15^2}}$

$\huge{\sf{64+15}}$

$\huge{\sf{\sqrt{64+225}}$

$\huge{\sf{a=17}</p><p>[tex] \huge{\sf{sin(A) =\frac{15}{17}}$

$\huge{\sf{sec(A)=\frac{17}{8}$