Answer:
E-F and E-D
C-B and C-D
Step-by-step explanation:
The circle and triangle are as shown. The options are:
- A-B and C-B
- E-F and E-D
- E-D and C-D
- A-F and E-F
- C-B and C-D
By drawing radius lines from the center of the circle to the tangent points B, D, and F, we can divide the triangle into 3 kites. Therefore, only segments that are legs of the same kite are congruent. So the answer must be E-F and E-D, and C-B and C-D.
Answer:
Option (3)
Step-by-step explanation:
From the figure attached,
AB and CD are two chords intersecting at O.
m∠AOD = 37°
m∠AOC + m∠AOD = 180° [Since these angles are supplementary angles]
m∠AOC = 180° - 37°
= 143°
By the theorem of intersecting chords,
Measure of angle formed is the half of the sum of measures of the arcs intercepted by the angle and vertical angle.
m∠AOC = 
143° = ![\frac{1}{2}[(x+5)+(x-5)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%28x%2B5%29%2B%28x-5%29%5D)
143° = x
Therefore, Option (3) will be the answer.
Let Xavier's favourite fraction be a/b, Yessie's favourite fraction = b/a and Zorro's favourite fraction = c/d,
c/d x a/b = 12/35 . . . . . . . . (1)
c/d x b/a = 15/7 . . . . . . . . (2)
(1) x (2) = c/d x a/b x c/d x b/a = 12/35 x 15/7
c^2 / d^2 = 36/49
c^2 = 36
c = 6
d^2 = 49
d = 7
Xaviers favourite fraction = 12/35 / 6/7 = 2/5
Yessies favourite fraction = 5/2
Zorro favourite fraction = 6/7
B.) 30 degrees would be correct here
