Question

if sin A = 8/17 then find cos A and Tan A​

Answer 1

GIVEN :–

• sin (A) = 8/17

TO FIND :–

• cos (A) = ? and tan (A) = ?

SOLUTION :–

$\\ \implies{ \bold{ \sin(A) = \dfrac{8}{17} }}$

• Using identity –

$\\ \longrightarrow \: { \bold{ \sin {}^{2} (A) + \cos {}^{2} (A) = 1 }}$

• So that –

$\\ \implies \: { \bold{ \cos {}^{2} (A) = 1 - \sin {}^{2} (A) }}$

$\\ \implies \: { \bold{ \cos(A) = \sqrt{1 - \sin {}^{2} (A) }}}$

• Now put the values –

$\\ \implies \: { \bold{ \cos(A) = \sqrt{1 - { \left(\dfrac{8}{17} \right)}^{2} }}}$

$\\ \implies \: { \bold{ \cos(A) = \sqrt{1 - { \left(\dfrac{64}{289} \right)} }}}$

$\\ \implies \: { \bold{ \cos(A) = \sqrt{ \left(\dfrac{289 - 64}{289} \right)} }}$

$\\ \implies \: { \bold{ \cos(A) = \sqrt{ \left(\dfrac{225}{289} \right)} }}$

$\\ \longrightarrow \: \large { \boxed{ \bold{ \cos(A) = \dfrac{15}{17} }}}$

• We know that –

$\\ \longrightarrow \: { \bold{ \tan(A) = \dfrac{ \sin(A) }{ \cos(A) } }}$

• So that –

$\\ \implies \: { \bold{ \tan(A) = \dfrac{ \dfrac{8}{ \cancel{17}} }{ \dfrac{15}{ \cancel{17}} } }}$

$\\ \longrightarrow \: \large { \boxed{ \bold{ \tan(A) = \dfrac{ 8 }{ 15 } }}}$

Answer 2

$\bigstar\Large\underline{\green{\rm{DIAGRAM}}}:$

$\setlength{\unitlength} {1mm}\begin{picture} (5,5)\put(0,0){\line(1,0) {30}}\put(30,0){\line(0,1) {40}}\put(0,0){\line(3,4) {30}}\put(13,-4){3\ cm} \put(32,17){4\ cm}\put(8,23) {5 cm}\put(3,1){ \theta } \end{picture}$

$\bigstar\Large\rm \underline{\purple{GIVEN:}}$

• sinA = $\sf\dfrac{8}{17}$

$\bigstar\Large\rm \underline{\orange{TO\quad FIND:}}$

• cosA
• tanA

$\bigstar\Large \underline{\green{\rm{SOLUTION:}}}$

Given ,

sinA = $\sf\dfrac{8}{17}[\tex]</p><p>That means, </p><p>Oppposite side of angle A = 8</p><p>Hypotenuse = 17</p><p></p><p>Finding adjacent side of angle A using Pythagoras theorem</p><p>(Hypotenuse)[tex]^2$ = (Base)$^2$ + (Height)$^2$

17$^2$ = 8$^2$ + H$^2$

289 - 64 = H$^2$

H = $\sf\sqrt{225}$

H = 15

Now,

★ cosA = $\sf\dfrac{Adjacent}{Hypotenuse}= \dfrac{15}{17}$

★ tanA = $\sf\dfrac {Opposite}{Hypotenuse} = \dfrac{8}{15}$