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3 years ago

OM , ON and OP are the angle bisectors of ∠AOB, ∠BOC, and ∠COD respectively. ∠AOD and ∠BOE are straight angles.

Mathematics

1 answer:

Orlov [11]3 years ago

3 0

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

$2x+2y+2z=180^\circ$

We are given that:

m∠MOP = 110°.

From the diagram

∠MOP=x+2y+z

Therefore:

x+2y+z=110°.

Solving simultaneously by subtraction

$2x+2y+2z=180^\circ$

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

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1. According to the plots, the curves intersect when $x=90^\circ$ and $x=360^\circ$ .

We can confirm this algebraically.

$\sin(x) + 2 = -\cos(x) + 3$

$\sin(x) + \cos(x) = 1$

$\sqrt2 \sin\left(x + 45^\circ\right) = 1$

$\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}$

$x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n$

(where $n$ is an integer)

$x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n$

$x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n$

We get the two solutions we found in the interval [0°, 360°] with $n=1$ in the first case, and $n=0$ in the second case.

2. We have $\sin(x)=0$ when $x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}$ . For the given plot domain [0°, 360°], this happens when $180^\circ < x < 360^\circ$ .