Question

1. in triangle abc, side bc = 14 yd, side ab = 13 yd, angle b = 109°

Explanation:

Step1: Apply Law of Cosines

We use the Law of Cosines to find side $AC$, since we know two sides and the included angle. The formula is $AC^2 = BC^2 + AB^2 - 2 \cdot BC \cdot AB \cdot \cos(\angle B)$
$AC^2 = 14^2 + 13^2 - 2 \cdot 14 \cdot 13 \cdot \cos(109^\circ)$

Step2: Calculate squared terms

Compute the squares of the known sides.
$14^2 = 196$, $13^2 = 169$
$AC^2 = 196 + 169 - 2 \cdot 14 \cdot 13 \cdot \cos(109^\circ)$

Step3: Compute product and cosine value

Calculate the product $2 \cdot 14 \cdot 13 = 364$, and $\cos(109^\circ) \approx -0.3256$
$AC^2 = 365 - 364 \cdot (-0.3256)$

Step4: Simplify the expression

Calculate the product and add to 365.
$AC^2 = 365 + 118.5184 = 483.5184$

Step5: Find square root

Take the square root to get $AC$.
$AC = \sqrt{483.5184}$

Answer:

$\approx 22.0$ yards