A right-angle triangle, with two sides adjacent to the right angle labeled x and y respectively, and the hypotenuse is labeled 12 times square root of 3. The angle between the hypotenuse and the side labeled x measures 30 degrees. Find the values of $x$ and $y$ . Write your answers in simplest form.

Respuesta :

Answer:

y = 6(sq rt 3)

x = 18

Step-by-step explanation:

y = (0.5)(12)(sq rt 3) = 6(sq rt 3)

x = (0.866)(12)(sq rt 3) = (10.392)(sq rt 3) = sq rt 324

reduced

x = 18

The trigonometric function gives the ratio of different sides of a right-angle triangle. The value of x and y is 18 and 6√3 units respectively.

What are Trigonometric functions?

The trigonometric function gives the ratio of different sides of a right-angle triangle.

[tex]\rm Sin \theta=\dfrac{Perpendicular}{Hypotenuse}\\\\\\Cos \theta=\dfrac{Base}{Hypotenuse}\\\\\\Tan \theta=\dfrac{Perpendicular}{Base}\\\\\\Cosec \theta=\dfrac{Hypotenuse}{Perpendicular}\\\\\\Sec \theta=\dfrac{Hypotenuse}{Base}\\\\\\Cot \theta=\dfrac{Base}{Perpendicular}\\\\\\[/tex]

where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.

Let the side with label x be the base of the triangle and y be the perpendicular.

As the length of the base is x while the angle is 30°, and the length of the hypotenuse is 12√3, therefore, the value of x can be written as,

[tex]Cos \theta=\dfrac{Base}{Hypotenuse}\\\\\\Cos (30^o)=\dfrac{x}{12\sqrt3}\\\\\\Cos (30^o) \times {12\sqrt3}={x}\\\\x = {12\sqrt3} \times \dfrac{{\sqrt3}}{2}\\\\x = 18[/tex]

Now, as we know that the length of the perpendicular side is y, therefore, the value of y can be written as,

[tex]\rm Sin \theta=\dfrac{Perpendicular}{Hypotenuse}\\\\\\Sin (30^o)=\dfrac{y}{12\sqrt3}\\\\\\Sin (30^o) \times {12\sqrt3} ={y}\\\\\\y = \dfrac{1}{2} \times 12\sqrt3\\\\\\y = 6\sqrt3[/tex]

Hence, the value of x and y is 18 and 6√3 units respectively.

Learn more about Trigonometric functions:

https://brainly.com/question/6904750

RELAXING NOICE
Relax