Question

problem 6. (openstax 5.4.42) find x in the following triangle: this is different than the problems we have covered in class! but try to think about how you can relate it back to what we covered in class! you will need a calculator for this problem.

Explanation:

Step1: Use the area formula for triangles

The area of a triangle can be expressed in two - part form. Let's assume the height corresponding to the side of length \(x\) is \(h\). The area \(A=\frac{1}{2}\times x\times h\). Also, using the other angles and side. We know that if we consider the two right - angled sub - triangles formed by the height. Let the height \(h_1\) of the upper right - angled triangle: \(h_1 = 82\sin63^{\circ}\), and the base of the upper right - angled triangle \(b_1=82\cos63^{\circ}\). Let the height \(h_2\) of the lower right - angled triangle: \(h_2 = 39\sin39^{\circ}\), and the base of the lower right - angled triangle \(b_2 = 39\cos39^{\circ}\). Since \(h_1 = h_2\) (the height of the whole triangle), and we can also use the formula \(A=\frac{1}{2}ab\sin C\). The area of the whole triangle can be written as \(A=\frac{1}{2}\times82\times39\times\sin(63 + 39)^{\circ}\). Also, \(A=\frac{1}{2}\times x\times82\sin63^{\circ}+\frac{1}{2}\times x\times39\sin39^{\circ}\).

Another way: Using the formula \(A=\frac{1}{2}ab\sin C\) for the whole triangle \(C = 63^{\circ}+39^{\circ}=102^{\circ}\), \(a = 82\), \(b = 39\). And also \(A=\frac{1}{2}x\times h\), where \(h\) is the height of the triangle with respect to the side \(x\). The height \(h\) can be calculated from the two sub - triangles.

We know that \(x=\frac{82\sin63^{\circ}+39\sin39^{\circ}}{\sin(180-(63 + 39))^{\circ}}\). Since \(\sin(180-(63 + 39))^{\circ}=\sin(63 + 39)^{\circ}=\sin102^{\circ}\approx0.9781\), \(\sin63^{\circ}\approx0.8910\), \(\sin39^{\circ}\approx0.6293\)

Step2: Calculate the value of \(x\)

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Answer:

\(x\approx99\)