In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.
(a) m ZVUX =
(b) m ZUXW =
Answers
Given:
In the circle, [tex]m(\overarc{VUX})=152^\circ[/tex] and [tex]m(\angle MUV)=77^\circ[/tex].
To find:
The following measures:
(a) [tex]m\angle VUX[/tex]
(b) [tex]m\angle UXW[/tex]
Solution:
According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.
[tex]m(VUX)=2\times m\angle VWX[/tex]
[tex]152^\circ=2\times m\angle VWX[/tex]
[tex]\dfrac{152^\circ}{2}=m\angle VWX[/tex]
[tex]76^\circ=m\angle VWX[/tex]
In a cyclic quadrilateral, the opposite angles are supplementary angles.
UVWX is a cyclic quadrilateral. So,
[tex]m\angle VUX+m\angle VWX=180^\circ[/tex] [Opposite angles of a cyclic quadrilateral]
[tex]m\angle VUX+76^\circ=180^\circ[/tex]
[tex]m\angle VUX=180^\circ-76^\circ[/tex]
[tex]m\angle VUX=104^\circ[/tex]
Now,
[tex]m\angle UXW+m\angle UVW=180^\circ[/tex] [Opposite angles of a cyclic quadrilateral]
[tex]m\angle UXW+77^\circ =180^\circ[/tex]
[tex]m\angle UXW=180^\circ-77^\circ[/tex]
[tex]m\angle UXW=103^\circ[/tex]
Therefore, [tex]m\angle VUX=104^\circ[/tex] and [tex]m\angle UXW=103^\circ[/tex] .
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Answers
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Answers
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HELP Use either law of sines or law of cosine. Need help on this problem! show work please!
Answers
Answer: x = 15.035677095729 approximately
Round this however you need to.
=================================================
Explanation:
I'm assuming you want to find the value of x, which your diagram is showing to be the length of segment QR.
If so, then we'll need to find the measure of angle Q first. Using the law of sines, we get the following:
sin(Q)/q = sin(R)/r
sin(Q)/PR = sin(R)/PQ
sin(Q)/13 = sin(85)/19
sin(Q) = 13*sin(85)/19
sin(Q) = 0.6816068987
Q = arcsin(0.6816068987) ... or ... Q = 180-arcsin(0.6816068987)
Q = 42.9693397461 ... or ... Q = 137.0306602539
These values are approximate.
----------------
Now if Q = 42.9693397461 approximately, then angle P is
P = 180-Q-R
P = 180-42.9693397461-85
P = 52.0306602539
Similarly, if Q = 137.0306602539 approximately, then,
P = 180-Q-R
P = 180-137.0306602539-85
P = -42.0306602539
A negative angle is not possible, so we'll ignore Q = 137.0306602539
----------------
The only possible value of angle P is approximately P = 52.0306602539
Let's apply the law of sines again to find side p, aka segment QR
sin(P)/p = sin(R)/r
sin(P)/QR = sin(R)/PQ
sin(52.0306602539)/x = sin(85)/19
19*sin(52.0306602539) = x*sin(85)
19*sin(52.0306602539)/sin(85) = x
x = 15.035677095729
This value is approximate.
Round this value however you need to.