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In 
DeltaGHI,g=9.8cm,i=9.6cm and 
/_I=43^(@). Find all possible values of 
/_G, to the nearest 10th of a degree.
Answer:

In ΔGHI,g=9.8 cm,i=9.6 cm \Delta \mathrm{GHI}, g=9.8 \mathrm{~cm}, i=9.6 \mathrm{~cm} and I=43 \angle \mathrm{I}=43^{\circ} . Find all possible values of G \angle \mathrm{G} , to the nearest 1010th of a degree.\newlineAnswer:

Full solution

Q. In ΔGHI,g=9.8 cm,i=9.6 cm \Delta \mathrm{GHI}, g=9.8 \mathrm{~cm}, i=9.6 \mathrm{~cm} and I=43 \angle \mathrm{I}=43^{\circ} . Find all possible values of G \angle \mathrm{G} , to the nearest 1010th of a degree.\newlineAnswer:
  1. Law of Sines Formula: To find the possible values of angle GG, 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:\newlineasin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\newlinewhere aa, bb, and cc are the lengths of the sides, and AA, BB, and CC are the opposite angles. In our case, we have:\newlinegsin(G)=isin(I)\frac{g}{\sin(G)} = \frac{i}{\sin(I)}\newlineWe can plug in the values we know to find sin(G)\sin(G):\newline9.8sin(G)=9.6sin(43 degrees)\frac{9.8}{\sin(G)} = \frac{9.6}{\sin(43 \text{ degrees})}\newlineFirst, we calculate sin(43 degrees)\sin(43 \text{ degrees}):\newlinesin(43 degrees)0.682\sin(43 \text{ degrees}) \approx 0.682\newlineNow we can solve for sin(G)\sin(G):\newlinesin(G)=9.8×sin(43 degrees)9.6\sin(G) = \frac{9.8 \times \sin(43 \text{ degrees})}{9.6}\newlinesin(G)9.8×0.6829.6\sin(G) \approx \frac{9.8 \times 0.682}{9.6}\newlinesin(G)6.68849.6\sin(G) \approx \frac{6.6884}{9.6}\newlinesin(G)0.6967\sin(G) \approx 0.6967
  2. Calculating sin(G): Now that we have the value of sin(G)\sin(G), we can find angle GG by taking the inverse sine (arcsin) of sin(G)\sin(G). However, since the sine function is positive in both the first and second quadrants, there are two possible angles for GG that have the same sine value. These angles are supplementary, meaning they add up to 180180 degrees. We will find the first angle, which is the acute angle:\newlineGarcsin(0.6967)G \approx \arcsin(0.6967)\newlineG44.4G \approx 44.4 degrees (to the nearest tenth)
  3. Finding Angle G: To find the second possible value for angle G, we subtract the first value from 180180 degrees:\newline180180 degrees - 44.444.4 degrees =135.6= 135.6 degrees\newlineSo the second possible value for angle G is 135.6135.6 degrees (to the nearest tenth).
  4. Checking Validity: We must now check if the second possible value for angle GG is valid by ensuring that the sum of angles in a triangle is 180180 degrees. We already have angle II as 4343 degrees, and we need to check if angle HH can be a positive acute angle when angle GG is 135.6135.6 degrees.\newlineSum of angles in a triangle = 180180 degrees\newlineAngle HH = 180180 degrees - Angle GG - Angle II\newlineAngle HH = 180180 degrees - 135.6135.6 degrees - 4343 degrees\newlineAngle HH = 18018077 degrees\newlineSince angle HH is positive and less than 180180 degrees, the second value for angle GG is valid.

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