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$In /_\EFG,e=680cm,g=940cm and /_G=21^(@). Find all possible values of /_E, to the nearest degree. Answer:$

In $\triangle \mathrm{EFG}, e=680 \mathrm{~cm}, g=940 \mathrm{~cm}$ and $\angle \mathrm{G}=21^{\circ}$ . Find all possible values of $\angle \mathrm{E}$ , to the nearest degree. $\newline$ Answer:

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Algebra 1

Evaluate variable expressions for number sequences

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Q. In

\triangle \mathrm{EFG}, e=680 \mathrm{~cm}, g=940 \mathrm{~cm}

and

\angle \mathrm{G}=21^{\circ}

. Find all possible values of

\angle \mathrm{E}

, to the nearest degree.

\newline

Answer:

Apply Law of Sines: To find the possible values of angle $E$ , 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 opposite angle is constant for all three sides and angles in 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 opposite angles. In our case, we have: $\newline$ $\frac{e}{\sin(E)} = \frac{g}{\sin(G)}$ $\newline$ First, we need to calculate $\sin(G)$ . $\newline$ $\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$ $0$
Calculate $\sin(G)$ : Using a calculator, we find that: $\sin(21 \text{ degrees}) \approx 0.3584$ Now we can set up the equation using the Law of Sines: $\frac{680}{\sin(E)} = \frac{940}{0.3584}$ Next, we solve for $\sin(E)$ . $\sin(E) = \frac{680 \times 0.3584}{940}$
Solve for $\sin(E)$ : Performing the calculation gives us: $\newline$ $\sin(E) \approx 0.2581$ $\newline$ Now we need to find the angle $E$ whose sine is approximately $0.2581$ . We use the inverse sine function (also known as arcsin) to find this angle. $\newline$ $E \approx \arcsin(0.2581)$
Find angle $E$ : Using a calculator, we find that: $\newline$ $E \approx \arcsin(0.2581) \approx 15$ degrees $\newline$ However, since the sum of angles in any triangle must be $180$ degrees, we must check if there is another possible value for angle $E$ . This is because the sine function has the same value for two different angles in the range of $0$ to $180$ degrees (one acute and one obtuse). The other angle would be $180$ degrees - $15$ degrees = $165$ degrees. $\newline$ We must check if this obtuse angle is possible by adding it to the given angle $G$ and seeing if the sum is less than $180$ degrees. $\newline$ $15$ degrees + $E \approx \arcsin(0.2581) \approx 15$ $2$ degrees = $E \approx \arcsin(0.2581) \approx 15$ $3$ degrees $\newline$ $165$ degrees + $E \approx \arcsin(0.2581) \approx 15$ $2$ degrees = $E \approx \arcsin(0.2581) \approx 15$ $6$ degrees $\newline$ The sum of the obtuse angle and angle $G$ exceeds $180$ degrees, which is not possible in a triangle. Therefore, the only possible value for angle $E$ is the acute angle.