Respuesta :
Part A:
To find angle D, we make use of the rule of Sines because we know the measure of the side opposite angle D and we also know the meansure another angle of the triangle with the measure of the side opposite the angle.
Thus, we find angle D as follows:
[tex] \frac{62}{\sin{D}} = \frac{51}{\sin{53^o}} \\ \\ \Rightarrow \sin{D}= \frac{62\sin{53^o}}{51}} \\ \\ = \frac{49.5154}{51} =0.9709 \\ \\ \Rightarrow D=\sin^{-1}{0.9709} \\ \\ =76.14^o=76^o8'[/tex]
Therefore, angle D is 76°8'
Part B:
To find angle E, recall that the sum of the angles in a triangle is 180°.
Thus, 53° + 76°8' + E = 180°
E = 180° - 53° - 76°8' = 50°52'
Therefore, angle E is 50°52'
Part C:
To find the measure of side e, we apply the cosine rule as follows:
[tex]e^2=51^2+62^2-2(51)(62)\cos{50^o52'} \\ \\ =2,601+3,844-2,098.85=3,991.25 \\ \\ \Rightarrow e=\sqrt{3,991.25}=63.18[/tex]
Therefore, the measure of side e is 63.2
Part D
To find angle G, we make use of the rule of Sines because we know the measure of the side opposite angle G and we also know the measure another angle of the triangle with the measure of the side opposite the angle.
Thus, we find angle G as follows:
[tex]\frac{80}{\sin{G}} = \frac{62}{\sin{49^o}} \\ \\ \Rightarrow \sin{G}= \frac{80\sin{49^o}}{62} \\ \\ = \frac{60.3768}{62} =0.9738 \\ \\ \Rightarrow G =\sin^{-1}{0.9738}=76.86^o=76^o52'[/tex]
Therefore, angle G is 76°52'
Part E:
To find angle H, recall that the sum of the angles in a triangle is 180°.
Thus, 49° + 76°52' + H = 180°
H = 180° - 49° - 76°52' = 54°14'
Therefore, angle H is 54°14'
Part F:
To find the measure of side a, we apply the cosine rule as follows:
[tex]a^2=80^2+62^2-2(80)(62)\cos{54^o14'} \\ \\ =6,400+3,844-5,798.10=4,445.90 \\ \\ \Rightarrow a=\sqrt{4,445.90}=66.68[/tex]
Therefore, the measure of side a is 66.7
Part G:
To find angle A, we make use of the rule of Sines because we know the measure of the side opposite angle A and we also know the measure another angle of the triangle with the measure of the side opposite the angle.
Thus, we find angle A as follows:
[tex]\frac{66.7}{\sin{A}} = \frac{30}{\sin{25^o54'}} \\ \\ \Rightarrow \sin{A}= \frac{66.7\sin{25^o54'}}{30} \\ \\ = \frac{29.1347}{30} =0.9712 \\ \\ \Rightarrow A =\sin^{-1}{0.9712}=76.21^o=76^o12'[/tex]
Therefore, angle A is 76°12'
Part H:
To find angle B, recall that the sum of the angles in a triangle is 180°.
Thus, 25°54' + 76°12' + B = 180°
B = 180° - 25°54' - 76°12' = 77°53'
Therefore, angle B is 77°53'
Part I:
To find the measure of side b, we make use of the rule of sines
Thus, we find side b as follows:
[tex]\frac{b}{\sin{77^o53'}} = \frac{30}{\sin{25^o54'}} \\ \\ \Rightarrow b= \frac{30\sin{77^o53'}}{\sin{25^o54'}} \\ \\ = \frac{29.3317}{0.4368} =67.15[/tex]
Therefore, the measure of side b is 67.2
To find angle D, we make use of the rule of Sines because we know the measure of the side opposite angle D and we also know the meansure another angle of the triangle with the measure of the side opposite the angle.
Thus, we find angle D as follows:
[tex] \frac{62}{\sin{D}} = \frac{51}{\sin{53^o}} \\ \\ \Rightarrow \sin{D}= \frac{62\sin{53^o}}{51}} \\ \\ = \frac{49.5154}{51} =0.9709 \\ \\ \Rightarrow D=\sin^{-1}{0.9709} \\ \\ =76.14^o=76^o8'[/tex]
Therefore, angle D is 76°8'
Part B:
To find angle E, recall that the sum of the angles in a triangle is 180°.
Thus, 53° + 76°8' + E = 180°
E = 180° - 53° - 76°8' = 50°52'
Therefore, angle E is 50°52'
Part C:
To find the measure of side e, we apply the cosine rule as follows:
[tex]e^2=51^2+62^2-2(51)(62)\cos{50^o52'} \\ \\ =2,601+3,844-2,098.85=3,991.25 \\ \\ \Rightarrow e=\sqrt{3,991.25}=63.18[/tex]
Therefore, the measure of side e is 63.2
Part D
To find angle G, we make use of the rule of Sines because we know the measure of the side opposite angle G and we also know the measure another angle of the triangle with the measure of the side opposite the angle.
Thus, we find angle G as follows:
[tex]\frac{80}{\sin{G}} = \frac{62}{\sin{49^o}} \\ \\ \Rightarrow \sin{G}= \frac{80\sin{49^o}}{62} \\ \\ = \frac{60.3768}{62} =0.9738 \\ \\ \Rightarrow G =\sin^{-1}{0.9738}=76.86^o=76^o52'[/tex]
Therefore, angle G is 76°52'
Part E:
To find angle H, recall that the sum of the angles in a triangle is 180°.
Thus, 49° + 76°52' + H = 180°
H = 180° - 49° - 76°52' = 54°14'
Therefore, angle H is 54°14'
Part F:
To find the measure of side a, we apply the cosine rule as follows:
[tex]a^2=80^2+62^2-2(80)(62)\cos{54^o14'} \\ \\ =6,400+3,844-5,798.10=4,445.90 \\ \\ \Rightarrow a=\sqrt{4,445.90}=66.68[/tex]
Therefore, the measure of side a is 66.7
Part G:
To find angle A, we make use of the rule of Sines because we know the measure of the side opposite angle A and we also know the measure another angle of the triangle with the measure of the side opposite the angle.
Thus, we find angle A as follows:
[tex]\frac{66.7}{\sin{A}} = \frac{30}{\sin{25^o54'}} \\ \\ \Rightarrow \sin{A}= \frac{66.7\sin{25^o54'}}{30} \\ \\ = \frac{29.1347}{30} =0.9712 \\ \\ \Rightarrow A =\sin^{-1}{0.9712}=76.21^o=76^o12'[/tex]
Therefore, angle A is 76°12'
Part H:
To find angle B, recall that the sum of the angles in a triangle is 180°.
Thus, 25°54' + 76°12' + B = 180°
B = 180° - 25°54' - 76°12' = 77°53'
Therefore, angle B is 77°53'
Part I:
To find the measure of side b, we make use of the rule of sines
Thus, we find side b as follows:
[tex]\frac{b}{\sin{77^o53'}} = \frac{30}{\sin{25^o54'}} \\ \\ \Rightarrow b= \frac{30\sin{77^o53'}}{\sin{25^o54'}} \\ \\ = \frac{29.3317}{0.4368} =67.15[/tex]
Therefore, the measure of side b is 67.2
