Answer:
A)82.02 mi
B) 18.7° SE
Step-by-step explanation:
From the image attached, we can see the angles and distance depicted as given in the question. Using parallel angles, we have been able to establish that the internal angle at egg island is 100°.
A) Thus, we can find the distance between the home port and forrest island using law of cosines which is that;
a² = b² + c² - 2bc Cos A
Thus, let the distance between the home port and forrest island be x.
So,
x² = 40² + 65² - 2(40 × 65)cos 100
x² = 1600 + 4225 - (2 × 2600 × -0.1736)
x² = 6727.72
x = √6727.72
x = 82.02 mi
B) To find the bearing from Forrest Island back to his home port, we will make use of law of sines which is that;
A/sinA = b/sinB = c/sinC
82.02/sin 100 = 40/sinθ
Cross multiply to get;
sinθ = (40 × sin 100)/82.02
sin θ = 0.4803
θ = sin^(-1) 0.4803
θ = 28.7°
From the diagram we can see that from parallel angles, 10° is part of the total angle θ.
Thus, the bearing from Forrest Island back to his home port is;
28.7 - 10 = 18.7° SE
Answer:
In a quadratic equation of the shape:
y = a*x^2 + b*x + c
we hate that the discriminant is equal to:
D = b^2 - 4*a*c
This thing appears in the Bhaskara's formula for the roots of the quadratic equation:
You can see that the determinant is inside a square root, this means that if D is smaller than zero we will have imaginary roots (the graph never touches the x-axis)
If D = 0, the square root term dissapear, and this implies that both roots of the equation are the same, this means that the graph touches the x axis in only one point, wich coincides with the minimum/maximum of the graph)
If D > 0 we have two different roots, so the graph touches the x-axis in two different points.