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Chapter 5: Problem 29

Write the law of sines in the special case of a right triangle.

Short Answer

Expert verified
The law of sines for a right triangle, where angle \(C\) is the right angle, can be written as: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = c\), where \(\sin{90^\circ} = 1\).

Step by step solution

01

Understand the Law of Sines

The law of sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all three sides. Mathematically, this is expressed as: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}\) Here, \(a, b, c\) are the side lengths of the triangle and \(A, B, C\) are the angles opposite to the sides \(a, b, c\), respectively.
02

Consider a Right Triangle

In the special case of a right triangle, one of the angles is necessarily 90 degrees. Without loss of generality, let's assume angle \(C\) is the right angle of the triangle (90 degrees). This means that side \(c\) is the hypotenuse of the triangle.
03

Apply the Law of Sines to a Right Triangle

Now, let's apply the law of sines to this right triangle with angle \(C\) being the right angle. \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}\) Since angle \(C\) is 90 degrees, we have \(\sin{C} = \sin{90^\circ} = 1\). This simplifies the law of sines as follows: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = c\) Now, the law of sines for a right triangle can be written as: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = c\), where \(\sin{90^\circ} = 1\). This is the law of sines in the special case of a right triangle.

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Evaluate the indicated expressions assuming that $$ \begin{array}{ll} \cos x=\frac{1}{3} & \text { and } \sin y=\frac{1}{4} \\ \sin u=\frac{2}{3} & \text { and } \cos v=\frac{1}{5} \end{array} $$ Assume also that \(x\) and \(u\) are in the interval \(\left(0, \frac{\pi}{2}\right),\) that \(y\) is in the interval \(\left(\frac{\pi}{2}, \pi\right),\) and that \(v\) is in the interval \(\left(-\frac{\pi}{2}, 0\right) .\) $$\tan (u-v)$$

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