First, we are given that the inscribed angle of arc CB which is angle D is equal to 65°. This is half of the measure of the arc which is equal to the measure of the central angle, ∠O.
m∠O = 2 (65°) = 130°
Also, the measure of the angles where the tangent lines and the radii meet are equal to 90°. The sum of the measures of the angle of a quadrilateral ACOB is equal to 360°.
m∠O + m∠C + m∠B + m∠A = 360°
Substituting the known values,
130° + 90° + 90° + m∠A = 360°
The value of m∠A is equal to 50°.
<em>Answer: 50°</em>
Answer: x=4
Step-by-step explanation:
The perimeter´s formula for this kind of polygons is
The triangle perimeter would be:
(x+2)+(x+3)+(x+3)
And the rectangle:
(x+2)+(x+2)+x+x
For them to have the same value, we should equal them:
(x+2)+(x+3)+(x+3)=(x+2)+(x+2)+x+x
Working with this:
3x+8=4x+4
x=4
Answer:
AFH, BFG, CDG, EFG
Step-by-step explanation:
Please check the point position with your question, if there is any difference... the way to find out a right angle triangle is just like the attached picture.
Answer:
B. -11 + 7
Step-by-step explanation:
Minus a negative translates to add.
-11 - (-7) = -11 + 7
Answer:
slope of line ax + by = c is -a/b
so, slope of 7x +2y =5 is -7/2
