Math Problem Statement

Calculate the length x. The triangle has angles 48°, 59°, and side lengths of 6m and 8m as shown in the diagram.

Solution

To calculate the length xx in this triangle, we can use the Sine Rule, which is suitable for non-right-angled triangles when we know two angles and a side (or two sides and an angle opposite one of them).

The Sine Rule states: asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} Where:

  • a,b,ca, b, c are the lengths of the sides,
  • A,B,CA, B, C are the angles opposite these sides.

Given:

  • Angle opposite the side of 6 m: 4848^\circ,
  • Angle opposite the side of xx: 5959^\circ,
  • Side opposite 4848^\circ: 8 m,
  • Side opposite 5959^\circ: xx.

Using the Sine Rule: xsin59=8sin48\frac{x}{\sin 59^\circ} = \frac{8}{\sin 48^\circ}

We can rearrange to solve for xx: x=8sin59sin48x = \frac{8 \cdot \sin 59^\circ}{\sin 48^\circ}

Now, let's compute this:

  • sin590.8572\sin 59^\circ \approx 0.8572,
  • sin480.7431\sin 48^\circ \approx 0.7431.

Substitute these values: x=80.85720.74316.85760.74319.23m.x = \frac{8 \cdot 0.8572}{0.7431} \approx \frac{6.8576}{0.7431} \approx 9.23 \, \text{m}.

So, the length of xx is approximately 9.23 m, rounded to two significant figures.

Would you like a more detailed explanation or help with any part of this?

Here are some questions for further practice:

  1. What is the Sine Rule and when should it be used?
  2. Can the Cosine Rule be used for this problem? Why or why not?
  3. How would you solve for an unknown angle in a triangle using the Sine Rule?
  4. What is the difference between using degrees and radians in trigonometric calculations?
  5. How do you calculate the area of a triangle given two sides and an included angle?

Tip: Always check if the triangle is right-angled; if it is, the simpler trigonometric ratios (like sine, cosine, and tangent) can be used!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Rule

Formulas

Sine Rule: a/sin(A) = b/sin(B) = c/sin(C)

Theorems

Sine Rule

Suitable Grade Level

Grades 9-11