a circle is inscribed in a quadrilateral LMNO with angles measuring 87, 98, 48 and 127. What are the measures of the minor arcs formed

Question

a circle is inscribed in a quadrilateral LMNO with angles measuring 87, 98, 48 and 127. What are the measures of the minor arcs formed

Answer 1

Let's draw a picture of this situation and label some extra points so we're on the same page.

 

 

 

1. Draw a circle inscribed inside a quadrilateral with vertices L, M, N, and O.

 

2. Label L 87°, M 98°, N 48°, and O 127°.

 

3. Since LMNO is circumscribed about the circle, its sides are tangent to the circle. Label the points of tangency A (between L and M), B (between M and N), C (between N and O), and D (between O and L).

 

4. Label the center of the circle P and draw radii from P to A, B, C, and D.

 

 

 

We're interested in finding the minor arcs' measures. Those are AB, BC, CD, and DA. Start out by looking at AB, which is formed by the two sides of LMNO that intersect at M. AB is inside a new quadrilateral, PAMB. Since PA and PB are radii drawn to points of tangency, <PAM and <PBM are right angles. So, since quadrilaterals' angles sum to 360°, <M and <APB must add up to be the other 180°. They are supplementary. Solve for <APB:

 

<M + <APB = 180°

98° + <APB = 180°

<APB = 82°

 

Since <APB is a central angle measuring 82°, the arc it intercepts (AB) has the same measure. So, arc AB = 82°.

 

 

 

We'll do something similar for each of the other minor arcs. Consider BC. Just like AB, it's inside a new quadrilateral, BPCN. <BPC and <N are supplementary. So:

 

<N + <BPC = 180°

48° + <BPC = 180°

<BPC = 132°

 

<BPC is a central angle, so its intercepted arc BC is also 132°.

 

 

 

Try CD and DA on your own. The same methodology will work!