Question

a ski resort has a lift from a to c, as shown in the figure below. angle dac measures 34°. mountain officials want to build a new ski lift from b to c. the base b will be 890 feet from a, and the new lift will be 1825 feet long. what will be the measure of angle b? round your answer to the nearest tenth of a degree.

Explanation:

Step1: Apply the Law of Sines

In \(\triangle ABC\), by the Law of Sines \(\frac{AC}{\sin B}=\frac{BC}{\sin\angle BAC}\). First, find \(\angle BAC = 180^{\circ}- 34^{\circ}=146^{\circ}\), \(BC = 1825\), \(AC\) is not needed. Let \(\angle B=x\). The Law of Sines formula is \(\frac{AC}{\sin x}=\frac{1825}{\sin146^{\circ}}\). But we can also write \(\sin x=\frac{AC\sin146^{\circ}}{1825}\). Wait, no, correct Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin B}\), we can rewrite for \(\sin B\) as \(\sin B=\frac{AC\sin\angle BAC}{BC}\). Wait, no, actually, in \(\triangle ABC\), \(\angle BAC = 146^{\circ}\), \(BC = 1825\), assume \(AC\) is opposite to \(\angle B\), \(BC\) is opposite to \(\angle BAC\). The Law of Sines: \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). Wait, no, standard Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let \(a = BC = 1825\), \(A=\angle BAC = 146^{\circ}\), \(b = AC\), \(B\) is the angle we want. Wait, no, actually, if we consider \(\triangle ABC\), \(\angle BAC = 146^{\circ}\), \(BC\) is opposite \(\angle BAC\), \(AC\) is opposite \(\angle B\). Wait, no, no. Wait, in \(\triangle ABC\), \(\angle BAC = 146^{\circ}\), side \(BC = 1825\), side \(AB = 890\). Wait, no, wait the Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin B}=\frac{AB}{\sin C}\). But we can use \(\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin B}\). Wait, no, actually, we can use \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin C}\), but no, we need \(\angle B\). Wait, no, correct approach: In \(\triangle ABC\), by the Law of Sines \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). Wait, no, let's start over. Let’s denote in \(\triangle ABC\), \(\angle BAC = 146^{\circ}\), \(BC = 1825\), \(AB = 890\). By the Law of Sines \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). Wait, no, the Law of Sines is \(\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin B}=\frac{AB}{\sin C}\). But we can write \(\sin B=\frac{AC\sin\angle BAC}{BC}\). But we don't know \(AC\). Wait, no, wait, in right - triangle \(ADC\) (assuming \(CD\perp AB\) extended), but no, we can just use the Law of Sines directly on \(\triangle ABC\). \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). Wait, no, the Law of Sines formula is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let \(a = BC = 1825\), \(A=\angle BAC = 146^{\circ}\), \(c = AB = 890\), \(C\) is the angle at \(C\), \(B\) is the angle at \(B\). Then \(\frac{1825}{\sin146^{\circ}}=\frac{890}{\sin C}\), but we need \(\angle B\). Wait, no, \(\angle B+\angle BAC+\angle C=180^{\circ}\), \(\angle C = 180^{\circ}-\angle B - 146^{\circ}=34^{\circ}-\angle B\). But another way: \(\frac{\sin B}{AC}=\frac{\sin146^{\circ}}{1825}\). Wait, no, actually, if we use the Law of Sines \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). Wait, no, let's use the correct formula. In \(\triangle ABC\), \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). But we can also use \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). Wait, no, we can write \(\sin B=\frac{AC\sin\angle BAC}{BC}\). But we can find \(AC\) from right - triangle \(ADC\) (assuming \(CD\perp AB\) extended). Let’s assume \(CD\perp AB\) (extended). In right - triangle \(ADC\), \(\sin34^{\circ}=\frac{CD}{AC}\), \(\cos34^{\circ}=\frac{AD}{AC}\). But in \(\triangle BDC\), using Pythagoras \(BC^{2}=CD^{2}+(AD + AB)^{2}\). But this is complex. The Law of Sines is better. In \(\triangle ABC\), \(\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin B}\). Let’s assume \(AC\) is found from right - triangle (but no, the Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin C}\). But \(\angle C=180^{\circ}-\angle B - 146^{\circ}\). No, the Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin C}\), \(\frac{1825}{\sin146^{\circ}}=\frac{890}{\sin C}\), \(\sin C=\frac{890\sin146^{\circ}}{1825}\). \(\sin146^{\circ}=\sin(180 - 34)^{\circ}=\sin34^{\circ}\approx0.5592\). \(\sin C=\frac{890\times0.5592}{1825}\approx\frac{497.688}{1825}\approx0.2727\). \(\angle C\approx15.8^{\circ}\). Then \(\angle B=180^{\circ}-\angle BAC-\angle C=180 - 146-15.8 = 18.2^{\circ}\).

Another way: Using the Law of Sines directly \(\frac{\sin B}{AC}=\frac{\sin\angle BAC}{BC}\). But also, in \(\triangle ABC\), \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin C}\). \(\angle BAC = 146^{\circ}\), \(BC = 1825\), \(AB = 890\). \(\sin C=\frac{AB\sin\angle BAC}{BC}=\frac{890\sin146^{\circ}}{1825}\). Since \(\sin146^{\circ}=\sin34^{\circ}\), \(\sin C=\frac{890\times0.5592}{1825}\approx0.2727\), \(\angle C\approx15.8^{\circ}\). Then \(\angle B = 180-(146 + 15.8)=18.2^{\circ}\)

Answer:

\(18.2^{\circ}\)