Question

a quadrilateral
consider the following quadrilateral abcd.
what is, to the nearest unit, the measure of angle abc?

Explanation:

Step1: Connect A to C, find AC²

Using the Law of Cosines in $\triangle ADC$:
$$AC^2 = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cdot \cos(\angle ADC)$$
$$AC^2 = 25^2 + 32^2 - 2 \cdot 25 \cdot 32 \cdot \cos(89^\circ)$$
$$AC^2 = 625 + 1024 - 1600 \cdot 0.01745$$
$$AC^2 = 1649 - 27.92 = 1621.08$$

Step2: Calculate AC length

$$AC = \sqrt{1621.08} \approx 40.26 \text{ cm}$$

Step3: Use Law of Cosines in $\triangle ABC$

Let $\angle ABC = \theta$, rearrange to solve for $\cos\theta$:
$$\cos\theta = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$
$$\cos\theta = \frac{30^2 + 18^2 - 1621.08}{2 \cdot 30 \cdot 18}$$
$$\cos\theta = \frac{900 + 324 - 1621.08}{1080}$$
$$\cos\theta = \frac{1224 - 1621.08}{1080} = \frac{-397.08}{1080} \approx -0.3677$$

Step4: Find $\theta$ using arccosine

$$\theta = \arccos(-0.3677) \approx 111.6^\circ$$

Answer:

112