Question

ANY HELP WOULD BE APPRECIATED


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 :

• sin x = 3/5
• cos y = 3/5
• ratio = 1 : 1

Explanation :

first, let's find out the hypotenuse of the triangle using the pythagoras theorem

• hypotenuse = √((perpendicular)^2 + (base)^2)
• OP = √((8)^2 + (6)^2)
• OP = 10

sinθ = opposite/hypotenuse

• sin x = 6/10
• sin x= 3*2/5*2
• sin x = 3/5

cosθ = adjacent/hypotenuse

• cos y = 6/10
• cos y = 3*2/5*2
• cos y = 3/5

thus, the ratio of the values of sin x° and cos y° would be

• 3/5 : 3/5
• 1 : 1

therefore, the values of sin x and cos y are in a ratio of 1 : 1 which means they're equal in nature.

Answer 2

Answer:

$\sin(x^\circ) = \dfrac{3}{5}$

$\cos(y^\circ) = \dfrac{3}{5}$

ratio: 1:1

$\sin(x^\circ) = \cos(y^\circ)$

Step-by-step explanation:

Let's find the hypotenuse using Pythagorean theorem:

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

Where

• c is hypotenuse,
• a and b are other sides.

Here,

a = 6 and b = 8.

Substitute the value:

$c^2 = 6^2+8^2$

$c^2 = 36 + 64$

$c^2 = 100$

Taking square root on both sides:

$c = \sqrt{100}$

$c = 10$

So, the hypotenuse PO is 10.

To find the values of $\sin(x^\circ)$ and $\cos(y^\circ)$, we use the given information about the lengths of the sides of the right triangles formed by the angles $x^\circ$ and $y^\circ$.

For angle $x^\circ$:

• Opposite side: $6$
• Adjacent side: $8$
• Hypotenuse: $\textsf{PO = 10}$

Using the definitions of $\sin$ in a right triangle:

$\sin(x^\circ) = \dfrac{\textsf{opposite}}{\textsf{hypotenuse}} = \dfrac{6}{10} = \dfrac{3}{5}$

For angle $y^\circ$:

• Opposite side: $8$
• Adjacent side: $6$
• Hypotenuse: $\textsf{PO} = 10$

Using the definitions of $\cos$ in a right triangle:

$\cos(y^\circ) = \dfrac{\textsf{adjacent}}{\textsf{hypotenuse}} = \dfrac{6}{10}= \dfrac{3}{5}$

Since both $\sin(x^\circ)$ and $\cos(y^\circ)$ have the same ratio of their opposite and hypotenuse sides, so, they are equal.

$\sin(x^\circ) = \cos(y^\circ)$

Therefore, $\sin(x^\circ)$ and $\cos(y^\circ)$ share a equal relationship.