Taking
and differentiating both sides with respect to
yields
Solving for the first derivative, we have
Differentiating again gives
Solving for the second derivative, we have
Now, when
and
, we have
Both of these problems will be solved in a similar way, but with different numbers. First, we set up an equation with the values given. Then, we solve. Lastly, we plug into the original expressions to solve for the angles.
[23] ABD = 42°, DBC = 35°
(4x - 2) + (3x + 2) = 77°
4x+ 3x + 2 - 2 = 77°
4x+ 3x= 77°
7x= 77°
x= 11°
-
ABD = (4x - 2) = (4(11°) - 2) = 44° - 2 = 42°
DBC = (3x + 2) = (3(11°) + 2) = 33° + 2 = 35°
[24] ABD = 62°, DBC = 78°
(4x - 8) + (4x + 8) = 140°
4x + 4x + 8 - 8 = 140°
4x + 4x = 140°
8x = 140°
8x = 140°
x = 17.5°
-
ABD = (4x - 8) = (4(17.5°) - 8) = 70° - 8° = 62°
DBC =(4x + 8) = (4(17.5°) + 8) = 70° + 8° = 78°
The answer to your question is 13.5
Area = length * width
Area = (n yds + 40 yds) * 30 yds
or
Area = (30 yds * n yards) + (30 yds * 40 yards)
This is if you add the area of the rides to the area of the arcade (the first parenthesis is the rides, the second is the arcade.)
The Law of Cosines features the 3 side lengths of a triangle, plus the measure of the angle opposite one of those sides.
We want angle x, which is opposite the side of length 39.
Then: a^2 = b^2 - 2ab cos C becomes 39^2 = 36^2 + 59^2 - 2(36)(59)cos x
or 1521 = 3481 + 1296 - 2(36)(59) cos x
Subtract (3481+1296) from both sides: 1521 - 4777 = -4248cos x
-3256 = -4248cos x
-3256
Then: cosx = --------------- = 0.766
-4248
Solving for x: x = arccos -0.766 = 0.698 radian, or 40 degrees (answer)