Question

1. from point c, both ends a and b of a bridge are visible. if ac = 21 m, bc = 33 m, and c = 50°, find ab, the length of the bridge to the nearest tenth of a meter.

Explanation:

Step1: Apply Law of Cosines

We use the Law of Cosines for $\triangle ABC$:
$$AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(C)$$

Step2: Substitute given values

Plug in $AC=21$, $BC=33$, $C=50^\circ$:
$$AB^2 = 21^2 + 33^2 - 2 \cdot 21 \cdot 33 \cdot \cos(50^\circ)$$
Calculate each term:
$21^2=441$, $33^2=1089$, $2 \cdot 21 \cdot 33=1386$, $\cos(50^\circ)\approx0.6428$
$$AB^2 = 441 + 1089 - 1386 \cdot 0.6428$$

Step3: Compute intermediate values

$$441+1089=1530$$
$$1386 \cdot 0.6428 \approx 891.92$$
$$AB^2 \approx 1530 - 891.92 = 638.08$$

Step4: Solve for AB

Take square root of both sides:
$$AB \approx \sqrt{638.08}$$

Answer:

$25.3$ m