Let AB extended intersect DC extended at point E
<span>We now have right triangle BEC with E = 90 degrees </span>
<span>For triangle BEC: </span>
<span>Exterior angle at E = 90 </span>
<span>Exterior angle at C = 148 (given) </span>
<span>Exterior angle of all polygons add up to 360 degrees </span>
<span>Exterior angle at B = 360−148−90 = 122 </span>
<span>So in quadrilateral ABCD </span>
<span>B = 122 </span>
<span>D = 360−44−148−122 = 46</span>
SAS only. The 3rd side cannot be proven.
Here is the answer to your question
None because it is impossible for any triangle to have more than one obtuse angle.
Answer:
The answer is x = 10.
Step-by-step explanation:
Use the pythagorean theorem. A would be 6, B would be 8, and C would be solved by the product of that.
