Question

2 fill in the blank 1 point a path in a park forms a triangle as shown. what is the measure of angle a? round to the nearest tenth of a degree. m∠a≈ type your answer... 3 fill in the blank 1 point what is the length of y? round to the nearest tenth. y≈ type your answer... 4 multiple choice 1 point in △lmn,m∠l=108∘,m∠m=25∘ and l=12. what is m? about 5.3 about 20.8 27 about 0.19

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). For the triangle in problem 2, we have \(\frac{135}{\sin A}=\frac{77}{\sin(180 - 119 - A)}\). But more straightforwardly, using the Law of Sines formula \(\frac{BC}{\sin A}=\frac{AC}{\sin B}\). First, find side \(AB\) using the Law of Cosines \(AB^{2}=77^{2}+135^{2}-2\times77\times135\times\cos(119^{\circ})\), but since we can use the Law of Sines \(\frac{135}{\sin A}=\frac{77}{\sin B}\) and \(A + B+119^{\circ}=180^{\circ}\), \(B = 61^{\circ}-A\). However, the correct Law of Sines application is \(\frac{135}{\sin A}=\frac{77}{\sin(180 - 119 - A)}\) simplifies to \(\frac{135}{\sin A}=\frac{77}{\sin(61 - A)}\). Cross - multiply: \(135\sin(61 - A)=77\sin A\). Using \(\sin(A - B)=\sin A\cos B-\cos A\sin B\), \(\sin(61 - A)=\sin61\cos A-\cos61\sin A\). So \(135(\sin61\cos A-\cos61\sin A)=77\sin A\). \(135\sin61\cos A=(77 + 135\cos61)\sin A\). \(\tan A=\frac{135\sin61}{77 + 135\cos61}\). Calculate \(\sin61\approx0.8746\), \(\cos61\approx0.4848\). \(135\times0.8746 = 118.071\), \(135\times0.4848=65.448\), \(77 + 65.448 = 142.448\). \(\tan A=\frac{118.071}{142.448}\approx0.8289\). \(A=\arctan(0.8289)\approx39.5^{\circ}\)

For problem 3: In \(\triangle XYZ\), using the Law of Sines \(\frac{y}{\sin32^{\circ}}=\frac{10}{\sin(180 - 32 - 32)}\). Since \(\angle Z=180 - 32 - 32 = 116^{\circ}\), \(\frac{y}{\sin32^{\circ}}=\frac{10}{\sin116^{\circ}}\). \(\sin116^{\circ}=\sin(90 + 26)=\cos26\approx0.8988\), \(\sin32^{\circ}\approx0.5299\). \(y=\frac{10\times\sin32^{\circ}}{\sin116^{\circ}}=\frac{10\times0.5299}{0.8988}\approx5.9\)

For problem 4: In \(\triangle LMN\), using the Law of Sines \(\frac{m}{\sin M}=\frac{l}{\sin L}\). Given \(l = 12\), \(m\angle L=108^{\circ}\), \(m\angle M = 25^{\circ}\). \(m=\frac{12\times\sin25^{\circ}}{\sin108^{\circ}}\). \(\sin25^{\circ}\approx0.4226\), \(\sin108^{\circ}\approx0.9511\). \(m=\frac{12\times0.4226}{0.9511}=\frac{5.0712}{0.9511}\approx5.3\)

Answer:

Problem 2: \(39.5\)
Problem 3: \(5.9\)
Problem 4: about \(5.3\)