In /_\FGH,g=170cm,f=960cm and /_F=19^(@). Find all possible values of /_G, to the nearest 1oth of a degree. Answer:

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In  /_\FGH,g=170cm,f=960cm and  /_F=19^(@). Find all possible values of  /_G, to the nearest 1oth of a degree. Answer:

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Apply Law of Sines: To find the possible values of /_G, 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 the same for all sides and angles in the triangle. The formula is: $$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} $$ Here, we have side g opposite /_G, side f opposite /_F, and we are given the values of g, f, and /_F.Write Formula with Given Values: First, let's write the Law of Sines formula using the given values: $$ \frac{f}{\sin(F)} = \frac{g}{\sin(G)} $$ $$ \frac{960}{\sin(19°)} = \frac{170}{\sin(G)} $$Solve for sin(G): Now, we solve for $\sin(G)$: $$ \sin(G) = \frac{170 \cdot \sin(19°)}{960} $$Calculate sin(G) Value: We calculate the value of $\sin(G)$ using a calculator: $$ \sin(G) ≈ \frac{170 \cdot 0.32557}{960} $$ $$ \sin(G) ≈ \frac{55.3469}{960} $$ $$ \sin(G) ≈ 0.05765 $$Find Angle G: Next, we find the angle G by taking the inverse sine (arcsin) of $\sin(G)$: $$ G ≈ \arcsin(0.05765) $$Check Validity of Second Value: Using a calculator, we find: $$ G ≈ 3.3° $$ However, since the sine function is positive in both the first and second quadrants, there could be another possible value for angle G in the second quadrant. This value would be: $$ G' = 180° - G $$ $$ G' = 180° - 3.3° $$ $$ G' ≈ 176.7° $$We must check if the second possible value for angle G is valid by ensuring the sum of the angles in the triangle does not exceed 180°. Since we already have /_F = 19°, we add this to our two possible values for /_G: $$ 19° + 3.3° + G'' \leq 180° $$ $$ 19° + 176.7° + G'' \leq 180° $$ The first equation is valid for any angle G'', but the second equation is not possible because 19° + 176.7° already exceeds 180°. Therefore, the only possible value for /_G is 3.3°.

In $\triangle \mathrm{FGH}, g=170 \mathrm{~cm}, f=960 \mathrm{~cm}$ and $\angle \mathrm{F}=19^{\circ}$ . Find all possible values of $\angle \mathrm{G}$ , to the nearest $10$ th of a degree. $\newline$ Answer:

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In △FGH,g=170 cm,f=960 cm \triangle \mathrm{FGH}, g=170 \mathrm{~cm}, f=960 \mathrm{~cm} △FGH,g=170 cm,f=960 cm and ∠F=19∘ \angle \mathrm{F}=19^{\circ} ∠F=19∘. Find all possible values of ∠G \angle \mathrm{G} ∠G, to the nearest 101010th of a degree.\newlineAnswer:

In $\triangle \mathrm{FGH}, g=170 \mathrm{~cm}, f=960 \mathrm{~cm}$ and $\angle \mathrm{F}=19^{\circ}$ . Find all possible values of $\angle \mathrm{G}$ , to the nearest $10$ th of a degree. $\newline$ Answer: