Find triangle area using sine

• Derive the sine area formula from one-half base times perpendicular height.
• Identify the included angle between the two side lengths used in the formula.
• Compute and interpret area with appropriate square units and reasonableness checks.

Build the idea

Worked example

1. Step 1

   Insert eight and eleven for the enclosing sides and thirty degrees for their included angle.

   $$K=\frac12(8)(11)\sin30^\circ$$

2. Step 2

   Sine thirty degrees equals one half. The product one half times eighty-eight times one half gives twenty-two.

   $$K=44\left(\frac12\right)=22$$

3. Step 3

   The height relative to the side of length eight is eleven times one half, or five point five. One half times base eight times height five point five also gives twenty-two.

   $$\frac12(8)(5.5)=22$$

• Using an angle that is not between the two side lengths in the product.

  Mark the angle physically enclosed by the chosen sides; that angle determines the perpendicular component used as height.

• Reporting centimeters instead of square centimeters.

  Area multiplies two length dimensions, so its units are squared even when sine itself has no units.

Find the area of a triangle with sides six meters and ten meters enclosing a one-hundred-twenty-degree angle.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Lesson overview

   Two sides and their included angle determine a triangle's area because sine extracts the perpendicular height from one side. Every triangle has area one half base times perpendicular height. When the height is missing, dropping an altitude creates a right triangle that relates the height to a known side and angle. Derive the sine area formula from one-half base times perpendicular height.

2. Part 2

   Visual model and equations

   If sides a and b enclose angle C, the perpendicular component of b is b sine C. Substituting that height into the standard area formula produces one half a b sine C. Use side a as a horizontal base and side b as a slanted edge making angle C with the base. Dropping a vertical altitude shows height b sine C. The height h equals b times sine C. The area K equals one half a b times sine C.

   • The height h equals b times sine C.
   • The area K equals one half a b times sine C.
3. Part 3

   Worked example

   Find the area of a triangle with sides eight centimeters and eleven centimeters enclosing a thirty-degree angle. Use the two given sides with their included angle in the sine area formula, evaluate the special-angle sine, and report square units. Insert eight and eleven for the enclosing sides and thirty degrees for their included angle. Sine thirty degrees equals one half. The product one half times eighty-eight times one half gives twenty-two. The height relative to the side of length eight is eleven times one half, or five point five. One half times base eight times height five point five also gives twenty-two. The triangle's area is twenty-two square centimeters. The derived height is shorter than the side of length eleven, as expected for a thirty-degree component, and the base-height calculation matches.

   • K equals one half times eight times eleven times sine thirty degrees.
   • K equals forty-four times one half, which is twenty-two.
   • One half times eight times five point five equals twenty-two.
4. Part 4

   Checks, practice, and scope

   The angle must be the one between the chosen sides. Using a nonincluded angle combines measurements from different right-triangle relationships and generally gives the wrong height. Using an angle that is not between the two side lengths in the product. Mark the angle physically enclosed by the chosen sides; that angle determines the perpendicular component used as height. Reporting centimeters instead of square centimeters. Area multiplies two length dimensions, so its units are squared even when sine itself has no units. Find the area of a triangle with sides six meters and ten meters enclosing a one-hundred-twenty-degree angle. Use one half times six times ten times sine one hundred twenty degrees, noting that sine one hundred twenty equals square root three over two. The exact area is fifteen square root three square meters, approximately twenty-five point ninety-eight square meters. Sine extracts the perpendicular height component from a known side. The formula uses two sides and the angle included between them. A base-height recalculation provides a geometric verification. The method assumes two sides and their included angle are known; other data combinations may require a different trigonometric law first. Calculator approximations should retain enough precision until the final area is rounded.

Scope and limitations

• The method assumes two sides and their included angle are known; other data combinations may require a different trigonometric law first.
• Calculator approximations should retain enough precision until the final area is rounded.