Question

A surveyor took the following measurements from two irregularly shaped pieces of land. Some of the lengths and angle measures are missing. Find all missing lengths and angle measures. Round lengths to the nearest tenth and angle measures to the nearest minute

Answer 1

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:

$\frac{62}{\sin{D}} = \frac{51}{\sin{53^o}} \\ \\ \Rightarrow \sin{D}= \frac{62\sin{53^o}}{51}} \\ \\ = \frac{49.5154}{51} =0.9709 \\ \\ \Rightarrow D=\sin^{-1}{0.9709} \\ \\ =76.14^o=76^o8'$

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:

$e^2=51^2+62^2-2(51)(62)\cos{50^o52'} \\ \\ =2,601+3,844-2,098.85=3,991.25 \\ \\ \Rightarrow e=\sqrt{3,991.25}=63.18$

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:

$\frac{80}{\sin{G}} = \frac{62}{\sin{49^o}} \\ \\ \Rightarrow \sin{G}= \frac{80\sin{49^o}}{62} \\ \\ = \frac{60.3768}{62} =0.9738 \\ \\ \Rightarrow G =\sin^{-1}{0.9738}=76.86^o=76^o52'$

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:

$a^2=80^2+62^2-2(80)(62)\cos{54^o14'} \\ \\ =6,400+3,844-5,798.10=4,445.90 \\ \\ \Rightarrow a=\sqrt{4,445.90}=66.68$

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:

$\frac{66.7}{\sin{A}} = \frac{30}{\sin{25^o54'}} \\ \\ \Rightarrow \sin{A}= \frac{66.7\sin{25^o54'}}{30} \\ \\ = \frac{29.1347}{30} =0.9712 \\ \\ \Rightarrow A =\sin^{-1}{0.9712}=76.21^o=76^o12'$

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:

$\frac{b}{\sin{77^o53'}} = \frac{30}{\sin{25^o54'}} \\ \\ \Rightarrow b= \frac{30\sin{77^o53'}}{\sin{25^o54'}} \\ \\ = \frac{29.3317}{0.4368} =67.15$

Therefore, the measure of side b is 67.2