Question

Use the image below to answer the following question. Find the value of sin x® and cos yº What relationship do the ratios of sin x® and cos yº share?

Answer 1

Ⲁⲛ⳽ⲱⲉⲅ:

$\qquad\hookrightarrow\quad \sf {sin\:x = \dfrac{5}{13} }$

$\qquad\hookrightarrow\quad \sf {cos\:y= \dfrac{5}{13} }$

$\quad\rule{300pt}{1.5pt}\quad$

Ⲋⲟⳑⳙⲧⳕⲟⲛ :

As, we are given a right angled triangle with two sides 12 and 5, we can find the third side using Pythagoras theorem ,i.e,

$\qquad\bull ~{\boxed{\mathfrak{(Hypotenuse)^2 = (base)^2 + (perpendicular )^2 }} }~\bull$

$\quad\dashrightarrow\quad \sf { OP^2 = 12^2 + 5 ^2}$

$\quad\dashrightarrow\quad \sf { OP^2 = 144 + 25}$

$\quad\dashrightarrow\quad \sf {OP = \sqrt{ 169} }$

$\quad\dashrightarrow\quad \sf {OP = 13 }$

Using trigonometric identities:

$\qquad\bull ~{\boxed{\mathfrak{ sin \theta = \dfrac{Perpendicular}{Hypotenuse} }} }~\bull$

$\qquad\bull ~{\boxed{{{\mathfrak{cos\theta = \dfrac{Base}{hypotenuse} }}}} }~\bull$

Ⲧⲏⲉⲅⲉ⳨ⲟⲅⲉ,

For sin x , we have ,

• Perpendicular = 5
• Hypotenuse = 13
• Base = 12

$\quad\dashrightarrow\quad \sf {sin \:x = \dfrac{perpendicular}{hypotenuse} }$

$\quad\dashrightarrow\quad {\pmb{\sf {sin\:x = \dfrac{5}{13} }}}$

‎ㅤ

For cos y , we have ,

• Perpendicular = 12
• Hypotenuse = 13
• Base = 5

$\quad\dashrightarrow\quad \sf {cos \:y = \dfrac{base}{hypotenuse} }$

$\quad\dashrightarrow\quad {\pmb{\sf {cos\:y= \dfrac{5}{13}}} }$

$\rule{300pt}{3pt}$

Answer 2

Answer:

sin(x) and cos(y) are equal and have ratio of 1 : 1

First find the missing hypotenuse using Pythagoras theorem:

a² + b² = c²

12² + 5² = c²

c = √144+25

c = 13

using sine rule:

$sin(x) = \dfrac{opposite}{hypotenuse}$

$sin(x) = \dfrac{5}{13}$

using cosine rule:

$cos(y) = \dfrac{adjacent}{hypotenuse}$

$cos(y) = \dfrac{5}{13}$