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A triangular building lot is at a intersection of 2 roads which meet at an angle of 108 degrees.
The frontage on one street is 15.8 meters and the frontage on the second street is 10.2 meters.
a) What is the area of the lot?
b) What is the length of the third side of the lot?
Can you please help me ? I don't get it , thanks so much in advance:)
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(a) To solve (a), use this formula for the triangle area
area = ,
where 'a' and 'b' are two given sides of the triangle and 'C' is the angle
between these sides. In your case it works this way
area = = = 75.181 m^2 (rounded).
(b) To solve (b), use the cosine law formula. It allows to calculate the side 'c'
of the triangle, when two other sides 'a' and 'b' are given and the angle 'C'
between 'a' and 'b' is known
c = = =
= = 21.29 meters (rounded).
At this point, the problem is solved completely.