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

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

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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:  a/sin(A) = b/sin(B) = c/sin(C)  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:  e/sin(E) = g/sin(G)  First, we need to calculate sin(G). sin(G) = sin(21 degrees)Calculate sin(G): Using a calculator, we find that: sin(21 degrees) ≈ 0.3584  Now we can set up the equation using the Law of Sines: 680/sin(E) = 940/0.3584  Next, we solve for sin(E). sin(E) = 680 * 0.3584 / 940Solve for sin(E): Performing the calculation gives us: sin(E) ≈ 0.2581  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. E ≈ arcsin(0.2581)Find angle E: Using a calculator, we find that: E ≈ arcsin(0.2581) ≈ 15 degrees  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.  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. 15 degrees + 21 degrees = 36 degrees 165 degrees + 21 degrees = 186 degrees  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.

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

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: