Question

try this #1: find side a.

find side a: \\(\frac{a}{\text{sine } a} = \frac{b}{\text{sine } b} = \frac{c}{\text{sine } c}\\)

Explanation:

🆕 New Concept Discovered: Law of Sines
Using ratios to find sides in non-right triangles

Step 1: Find the missing angle \( C \)

The sum of all interior angles in any triangle is always \( 180^\circ \). We are given:

• \( A = 30^\circ \)
• \( B = 95^\circ \)

We can find angle \( C \) by subtracting these from \( 180^\circ \):

$$ C = 180^\circ - 30^\circ - 95^\circ $$

$$ C = 55^\circ $$

Step 2: Set up the Law of Sines

The Law of Sines relates the lengths of the sides of a triangle to the sines of its angles:

$$ \frac{a}{\sin A} = \frac{c}{\sin C} $$

We know:

• \( A = 30^\circ \)
• \( c = 10\text{ ft} \)
• \( C = 55^\circ \)

Substitute these values into the formula:

$$ \frac{a}{\sin 30^\circ} = \frac{10}{\sin 55^\circ} $$

Step 3: Solve for side \( a \)

Isolate \( a \) by multiplying both sides by \( \sin 30^\circ \):

$$ a = \frac{10 \cdot \sin 30^\circ}{\sin 55^\circ} $$

Using the known value \( \sin 30^\circ = 0.5 \) and approximating \( \sin 55^\circ \approx 0.8192 \):

$$ a = \frac{10 \cdot 0.5}{0.8192} $$

$$ a = \frac{5}{0.8192} $$

$$ a \approx 6.10\text{ ft} $$

Answer:

$$ a \approx 6.10\text{ ft} $$