Question

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

Answer 1

Answer:

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

• $\theta$ is the angle
• O is the side opposite the angle
• A is the side adjacent the angle
• H is the hypotenuse (the side opposite the right angle)

To find sin x and cos y, we must first find the length of the hypotenuse (PO) of the given right triangle.  

To do this, use Pythagoras' Theorem:

$a^2+b^2=c^2 \quad \textsf{(where a and b are the legs, and c is the hypotenuse)}$

Given:

• a = 6
• b = 8
• c = PO

$\implies 6^2+8^2=PO^2$

$\implies PO^2=100$

$\implies PO=\sqrt{100}$

$\implies PO=10$

Using the found value of the hypotenuse and the trig ratios quoted above:

$\implies \sf \sin(x)=\dfrac{6}{10}=\dfrac{3}{5}$

$\implies \sf \cos(y)=\dfrac{6}{10}=\dfrac{3}{5}$

Therefore, $\sf \sin(x)=\cos(y)$

Answer 2

Find Hypotenuse

• √6²+8²
• √10²
• 10

$\\ \rm\Rrightarrow sinx=\dfrac{Perpendicular}{Hypotenuse}$

$\\ \rm\Rrightarrow sinx=\dfrac{6}{10}$

$\\ \rm\Rrightarrow sinx=\dfrac{3}{5}$

And

$\\ \rm\Rrightarrow cosy=\dfrac{Base}{Hypotenuse}$

$\\ \rm\Rrightarrow cosy=\dfrac{6}{10}$

$\\ \rm\Rrightarrow cosy=\dfrac{3}{5}$