Question

Given 15 cotA=8,find sinA and secA​

Answer 1

${\boxed {\tt{GIVEN:-}}}$

$\tt{15 \cot A = 8}$

${\boxed {\tt{FIND:-}}}$

$\tt{ \sin A \: and \: \sec A}$

${\boxed {\tt{SOLUTION:-}}}$

$\tt{draw \: a \: right \: \triangle ABC \: in \: which \: \angle B = 90 \degree}$

$\tt{given \: in \: attachment}$

$\tt{now \: we \: have}$

$\tt{15 \cot A = 8 \: \: \: \: \cot A = \frac{8}{15} = \frac{AB}{BC} }$

$\tt{ = \frac{base}{perpendicular} }$

$\tt {let \: AB = 8k \: and \: BC = 15k}$

$\tt then \: AC = \sqrt{(AB {)}^{2} + ( {BC)}^{2} } \: \: \: (pythogoras \: theorem)$

$\tt AC = \sqrt{(8k {)}^{2} + ( {15k)}^{2} } \\ = \tt \sqrt{ {64k}^{2} + {225}^{2} } = \sqrt{ {289k}^{2} }$

$\tt = 17k$

$\tt \sin A = \frac{perpendicular}{hypotenuse} = \frac{BC} {AC}= \frac{15 \cancel k}{17\cancel k} = \frac{15} {17}$

$\tt \sec A = \frac{hypotenuse}{base} = \frac{AC}{ AB}= \frac{17 \cancel k}{8\cancel k} = \frac{17} {8}$

Answer 2

Answer:

I hope get your ans mark me in brainly. . . . . . .