Answer:
1200
0.65
The answers are in the function
Answer:
74
Step-by-step explanation:
Say that arc JL going through M is arc E and JL going the other way is arc D
For the angle formed by two tangents, K=(1/2)(E-D)
64=E-D
Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148
Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212
E is then 212
64=212-D
212-64=D=148
Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74
That would be A I believe
Answer:C
Step-by-step explanation:
Answer:
In quadrilateral ABCD we have
AC = AD
and AB being the bisector of ∠A.
Now, in ΔABC and ΔABD,
AC = AD
[Given]
AB = AB
[Common]
∠CAB = ∠DAB [∴ AB bisects ∠CAD]
∴ Using SAS criteria, we have
ΔABC ≌ ΔABD.
∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.
∴ BC = BD.
