Answer:
BD = <u>1</u> unit
AD = <u>1</u> unit
AB = <u>1.6</u> units
AC = <u>1.6</u> units
Step-by-step explanation:
In the picture attached, the triangle ABC is shown.
Given that triangle ABC is isosceles, then ∠B = ∠C
∠A + ∠B + ∠C = 180°
36° + 2∠B = 180°
∠B = (180° - 36°)/2
∠B = ∠C = 72°
From Law of Sines:
sin(∠A)/BC = sin(∠B)/AC = sin(∠C)/AB
(Remember that BC is 1 unit long)
AB = AC = sin(72°)/sin(36°) = 1.6
In triangle ABD, ∠B = 72°/2 = 36°, then:
∠A + ∠B + ∠D = 180°
36° + 36° + ∠D = 180°
∠D = 180° - 36° - 36° = 108°
From Law of Sines:
sin(∠A)/BD = sin(∠B)/AD = sin(∠D)/AB
(now ∠A = ∠B)
BD = AD = sin(∠A)*AB/sin(∠D)
BD = AD = sin(36°)*1.6/sin(108°) = 1
P= M÷T , Price equals money raised divided by number of tickets sold
58 212 2800 900
/ \ / \ / \ / \
29 2 2 106 700 4 3 300
/ \ / \ / \ / \
2 53 70 10 2 2 3 100
/ \ / \ / \
7 10 5 2 2 50
/ \ / \
5 2 5 10
/ \
5 2
The prime factors are in bold.
Answer:
A 90
Step-by-step explanation:
multiple ways to prove this.
e.g. since the angle between the two lines from the center of the circle to the 2 tangent touching points is 90 degrees (that is the meaning of these 90 degrees here as the angle of the circle segment defined by the 2 tangent touching points and the circle center), the tangents have the same "behavior" as tan and cot = the tangents at the norm circle at 0 and 90 degrees. they hit each other outside of the circle again at 90 degrees.
another way
imagine the two right triangles of the tangents crossing point to the circle center and the tangent/circle touching points.
the Hypotenuse of each triangle is cutting the 90 degree angle of the circle segment exactly in half (due to the symmetry principle). so the angle between radius side and Hypotenuse is 90/2 = 45 degrees.
that means also the angle of such a right triangle at the tangent crossing point is 45 degrees (as the sum of all angles in a triangle must be 180, we have the remainder of 180 - 90 - 45 = 45 degrees).
the angles of both right triangles at that point are the same, and so we can add 45+45 = 90 degrees for the total angle at the tangent crossing point.
