Y=2-x^2
Vertex ( 0, 2 )
Focus : ( 0, 7/4 )
Axis of symmetry : x = 0
Directrix : y = 9/4
Vertex ( 0, 2 )
Focus : ( 0, 7/4 )
Axis of symmetry : x = 0
Directrix : y = 9/4
State the triangle congruence theorem that can be used to prove triangles congruent.
1 answer:
You might be interested in
Answer:
The answer is;
54.4
Step-by-step explanation:
The given parameter are;
Triangle ΔFGH = Right triangle
The length of segment = 48
The measure of ∠HFG = 28°
The measurement required = The measure of segment = x
The measure of ∠FHG = 90° angle opposite hypotenuse side of a right triangle
Therefore, ∠FGH = 180° - ∠HFG - ∠FHG = 180° - 90° - 28° = 62°
∠FGH = 62°
By sine rule, we have;
/(sin ∠FGH) =
/(sin(∠FHG)
By substituting the known values, we have;
48/(sin 62°) = x/(sin(90°)
sin(90°) = 1, therefore, we have;
x/1 = x = 48/(sin 62°) = 54.4 (by rounding the answer to the nearest tenth)
x = 54.4