Help algebra please??
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Find m∠BOC, if m∠MOP = 110°.
Answer:
m∠BOC= 40 degrees
Step-by-step explanation:
A diagram has been drawn and attached below.
- OM bisects AOB into angles x and x respectively
- ON bisects ∠BOC into angles y and y respectively
- OP bisects ∠COD into angles z and z respectively.
Since ∠AOD is a straight line
x+x+y+y+z+z=180 degrees
We are given that:
m∠MOP = 110°.
From the diagram
∠MOP=x+2y+z
Therefore:
x+2y+z=110°.
Solving simultaneously by subtraction
x+2y+z=110°.
We obtain:
x+z=70°
Since we are required to find ∠BOC
∠BOC=2y
Therefore from x+2y+z=110° (since x+z=70°)
70+2y=110
2y=110-70
2y=40
Therefore:
m∠BOC= 40 degrees
Answer:
Step-by-step explanation:
We khow that the equation of a circle is written this way :
(x-a)²+(y-b)²=r² where (x,y) are the coordinates of the cercle's points and (a,b) the coordinates of the cercle's center and r the radius .
Our task is to khow the values of a and b :
- We khow that the center is lying on the line 3x+2y=16⇒ 2y=-3x+16⇒ y=
x+8
- We khow that the points P and Q are two points in the cercle
- Let Ω be the center of this cercle
- we can notice that : PΩ AND QΩ are both equal to the radius ⇒ PΩ=QΩ= r
- So let's write the expression of this distance using vectors KHOWING THAT Ω(a,b)
- Vector PΩ(a-4,b-6) and Vector QΩ(a-8,b-2)
- PΩ=
and QΩ=
- Let's substitute a by x and b by y
- PΩ=QΩ we substitute each distance by its expression
- After simplyfying the expressions we get finally : -12+8x-8y=0
- now we have -12x +8x-8y=0 and the line equation 3x+2y-16=0
- these are simultanious equations so after solving them we get x=3.8 wich is approximatively 4 and y=2
- we substitute a by 4 and y by 2 in PΩ to get the radius
- we get r =
= 4
- so r²= 16
- then the equation is : (x-4)²+(y-2)²=16