Question

In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

Answer 1

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$.

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

$152^\circ=2\times m\angle VWX$

$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$          [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

Now,

$m\angle UXW+m\angle UVW=180^\circ$          [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

$m\angle UXW=180^\circ-77^\circ$

$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$  and $m\angle UXW=103^\circ$ .