Question

jenny and joe are standing on a riverbank, 220 meters apart, at points a and b respectively. (see the figure below ) joe is 250 meters from a house located across the river at point c. suppose that angle a (angle bac) is 50°. what is the measure of angle b (angle abc)? round your answer to the nearest tenth of a degree.

Explanation:

Step1: Identify triangle sides/angles

In $\triangle ABC$: $AB=220\ \text{m}$, $BC=250\ \text{m}$, $\angle BAC=50^\circ$, find $\angle ABC$.

Step2: Apply Law of Sines

Law of Sines: $\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin\angle ACB}$
Substitute values: $\frac{250}{\sin50^\circ}=\frac{220}{\sin\angle ACB}$

Step3: Solve for $\sin\angle ACB$

Rearrange: $\sin\angle ACB=\frac{220\times\sin50^\circ}{250}$
Calculate: $\sin50^\circ\approx0.7660$, so $\sin\angle ACB\approx\frac{220\times0.7660}{250}\approx0.6739$
Find $\angle ACB\approx\arcsin(0.6739)\approx42.4^\circ$

Step4: Calculate $\angle ABC$

Triangle angle sum: $\angle ABC=180^\circ-\angle BAC-\angle ACB$
Substitute: $\angle ABC\approx180^\circ-50^\circ-42.4^\circ$

Answer:

$87.6^\circ$