Question

brian wants to fence in his triangular plot of farm land that measures 1.1 by 1.5 by 2.2 miles. determine the angles at which the fences of the three sides will meet.
triangle image with sides: a to b is 1.5 mi, a to c is 1.1 mi, b to c is 2.2 mi
rounding each angle to the nearest degree:
$m\angle a \approx 115$ degrees (complete)
$m\angle b \approx \square$ degrees (done)

Explanation:

Step1: Identify sides for Law of Sines

In triangle \(ABC\), sides: \(AC = 1.1\), \(AB = 1.5\), \(BC = 2.2\), \(\angle A = 115^\circ\). Use Law of Sines: \(\frac{\sin B}{AC}=\frac{\sin A}{BC}\).

Step2: Substitute values

\(\frac{\sin B}{1.1}=\frac{\sin 115^\circ}{2.2}\). Simplify: \(\sin B=\frac{1.1\times\sin 115^\circ}{2.2}=\frac{\sin 115^\circ}{2}\).

Step3: Calculate \(\sin 115^\circ\)

\(\sin 115^\circ\approx\sin(90^\circ + 25^\circ)=\cos 25^\circ\approx0.9063\). Then \(\sin B\approx\frac{0.9063}{2}\approx0.4531\).

Step4: Find \(\angle B\)

\(\angle B\approx\arcsin(0.4531)\approx27^\circ\) (since \(\angle B\) is acute, as \(BC\) is the longest side, so \(\angle A\) is obtuse, others acute).

Answer:

\(27\)