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

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

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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: $$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} $$ 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: $$ r = 86 \text{ cm} $$ $$ q = 66 \text{ cm} $$ $$ Q = 5^\circ $$ We want to find R, so we will use the Law of Sines to set up our equation: $$ \frac{q}{\sin(Q)} = \frac{r}{\sin(R)} $$Set Up Equation: Substitute the known values into the equation: $$ \frac{66}{\sin(5^\circ)} = \frac{86}{\sin(R)} $$ Now, we solve for $\sin(R)$: $$ \sin(R) = \frac{86 \cdot \sin(5^\circ)}{66} $$Substitute Values: Calculate the value of $\sin(R)$ using a calculator: $$ \sin(R) = \frac{86 \cdot \sin(5^\circ)}{66} \approx 0.1085 $$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: $$ R \approx \arcsin(0.1085) $$Find Angle R: Calculate the value of R using a calculator: $$ R \approx \arcsin(0.1085) $$ 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: $$ R \approx \arcsin(0.1085) \approx 6^\circ $$ (rounded to the nearest degree)Calculate Supplementary Angle: Next, calculate the supplementary angle: $$ R' = 180^\circ - R \approx 180^\circ - 6^\circ = 174^\circ $$ (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 179 degrees, leaving only 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.

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

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: