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$In /_\QRS,r=86cm,q=66cm and /_Q=5^(@). Find all possible values of /_R, to the nearest degree. Answer:$

In $\triangle \mathrm{QRS}, r=86 \mathrm{~cm}, q=66 \mathrm{~cm}$ and $\angle \mathrm{Q}=5^{\circ}$ . Find all possible values of $\angle \mathrm{R}$ , to the nearest degree. $\newline$ Answer:

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Algebra 2

Complex conjugate theorem

Full solution

Q. In

\triangle \mathrm{QRS}, r=86 \mathrm{~cm}, q=66 \mathrm{~cm}

and

\angle \mathrm{Q}=5^{\circ}

. Find all possible values of

\angle \mathrm{R}

, to the nearest degree.

\newline

Answer:

Law of Sines: To find the possible values of /_R, we can use the Law of Sines, which states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle. The formula is: $\newline$ $\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$ $\newline$ where a, b, and c are the lengths of the sides, and A, B, and C are the angles opposite those sides, respectively. In this case, we have side r opposite angle R, side q opposite angle Q, and we need to find angle R.
Given Information: First, let's write down what we know: $\newline$ $r = 86 \text{ cm}$ $\newline$ $q = 66 \text{ cm}$ $\newline$ $Q = 5^\circ$ $\newline$ We want to find R, so we will use the Law of Sines to set up our equation: $\newline$ $\frac{q}{\sin(Q)} = \frac{r}{\sin(R)}$
Set Up Equation: Substitute the known values into the equation: $\newline$ $\frac{66}{\sin(5^\circ)} = \frac{86}{\sin(R)}$ $\newline$ Now, we solve for $\sin(R)$ : $\newline$ $\sin(R) = \frac{86 \cdot \sin(5^\circ)}{66}$
Substitute Values: Calculate the value of $\sin(R)$ using a calculator: $\newline$ $\sin(R) = \frac{86 \cdot \sin(5^\circ)}{66} \approx 0.1085$
Calculate Sin(R): Now, we need to find the angle R whose sine is approximately $0$ . $1085$ . We use the inverse sine function (also known as arcsin) to find the angle: $\newline$ $R \approx \arcsin(0.1085)$
Find Angle R: Calculate the value of R using a calculator: $\newline$ $R \approx \arcsin(0.1085)$ $\newline$ However, we must remember that the sine function has two possible angles that have the same sine value in the range of $0$ to $180$ degrees (since we are dealing with a triangle, the angle must be less than $180$ degrees). These are the acute angle and its supplementary angle. So we must consider both possibilities.
Calculate Acute Angle: First, calculate the acute angle: $\newline$ $R \approx \arcsin(0.1085) \approx 6^\circ$ $\newline$ (rounded to the nearest degree)
Calculate Supplementary Angle: Next, calculate the supplementary angle: $\newline$ $R' = 180^\circ - R \approx 180^\circ - 6^\circ = 174^\circ$ $\newline$ (rounded to the nearest degree)
Check Validity: We now have two possible values for angle $R$ . However, we must check if both are valid within the context of a triangle. The sum of angles in any triangle is $180$ degrees. We already have one angle $Q$ which is $5$ degrees. If we add the acute angle $R$ ( $6$ degrees), the sum is $11$ degrees, leaving $169$ degrees for the third angle, which is possible. But if we add the supplementary angle $R'$ ( $174$ degrees), the sum is $180$ $0$ degrees, leaving only $180$ $1$ degree for the third angle, which is not possible because the given side lengths would not form a triangle with such a small angle. Therefore, the only valid solution for angle $R$ is the acute angle.