Math Problem Statement

Solve for x in a right triangleThank you where for uploading the file. Please the let hyp me knowotenuse is how I can assist you with it!12 cm, the angle opposite the unknown side x is 62°, and the given triangle is labeled with the hypotenuse and angle.

Solution

To solve for xx in the given right triangle, we can use trigonometric functions. Let's break this down:

Step 1: Identify the known values

  • The hypotenuse (cc) = 12cm12 \, \text{cm}
  • The angle = 6262^\circ
  • The side we need to find is xx, which is the opposite side relative to the 6262^\circ angle.

Step 2: Use the sine function

The sine function relates the opposite side and the hypotenuse: sin(angle)=oppositehypotenuse\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}

Substituting the values: sin(62)=x12\sin(62^\circ) = \frac{x}{12}

Step 3: Solve for xx

x=12sin(62)x = 12 \cdot \sin(62^\circ)

Using a calculator: sin(62)0.8829\sin(62^\circ) \approx 0.8829

So: x=120.8829=10.5948x = 12 \cdot 0.8829 = 10.5948

Step 4: Round to the nearest hundredth

x10.59cmx \approx 10.59 \, \text{cm}

Final Answer:

x=10.59cmx = 10.59 \, \text{cm}


Would you like me to explain this further or go into more detail?

Related Questions:

  1. How do you use other trigonometric ratios (cosine, tangent) to solve for sides in a triangle?
  2. How can you solve for the adjacent side using the same diagram?
  3. What is the relationship between all three sides of a right triangle (Pythagorean theorem)?
  4. How do you calculate angles in a triangle if all sides are known?
  5. What are the differences between sine, cosine, and tangent in solving triangles?

Tip:

Always label the sides of a triangle (opposite, adjacent, hypotenuse) relative to the angle you’re working with. This helps you decide which trigonometric ratio to use!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Solve Right TriangleSine Function

Formulas

sin(angle) = opposite / hypotenuse

Theorems

Trigonometric Ratios in Right Triangles

Suitable Grade Level

Grades 8-10