SOLUTION: triangle ABC is inscribed in a circle. given that AB is a 40 degree arc and ABC is a 50 degree angle, find the sizes of the other arcs and angles in the figure. thankyou:]

Algebra ->  Algebra  -> Triangles -> SOLUTION: triangle ABC is inscribed in a circle. given that AB is a 40 degree arc and ABC is a 50 degree angle, find the sizes of the other arcs and angles in the figure. thankyou:]      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 148657: triangle ABC is inscribed in a circle. given that AB is a 40 degree arc and ABC is a 50 degree angle, find the sizes of the other arcs and angles in the figure.
thankyou:]
Answer by Edwin McCravy(6935) About Me  (Show Source):
You can put this solution on YOUR website!
Triangle ABC is inscribed in a circle. given that AB is a 40 degree arc and ABC is a 50 degree angle, find the sizes of the other arcs and angles in the figure 

drawing%28400%2C400%2C-1.5%2C1.5%2C-1.5%2C1.5%2C+circle%280%2C0%2C1%29%2C%0D%0Atriangle%28-1%2C0%2C-.7660444431%2C.6427876097%2C.17364818%2C-.9848078%29%2C%0D%0Alocate%28-1.1%2C.4%2C%2740%B0%27%29%2C+locate%28-.82%2C.45%2C%2750%B0%27%29%2C+locate%28-1.1%2C.1%2C%27A%27%29%2C%0D%0Alocate%28-.9%2C.7%2C%27B%27%29%2Clocate%28.2%2C-1%2C%27C%27%29+%29

Angle ACB is an inscribed angle, subtending a 40° arc.
An inscribed angle has 1%2F2 the measure of its inscribed arc.
Therefore angle ACB has measure 20°, so we write that in:

drawing%28400%2C400%2C-1.5%2C1.5%2C-1.5%2C1.5%2C+circle%280%2C0%2C1%29%2C%0D%0Atriangle%28-1%2C0%2C-.7660444431%2C.6427876097%2C.17364818%2C-.9848078%29%2C%0D%0Alocate%28-1.1%2C.4%2C%2740%B0%27%29%2C+locate%28-.82%2C.45%2C%2750%B0%27%29%2C+locate%28-1.1%2C.1%2C%27A%27%29%2C%0D%0Alocate%28-.9%2C.7%2C%27B%27%29%2Clocate%28.2%2C-1%2C%27C%27%29%2Clocate%28-.27%2C-.5%2C%2720%B0%27%29+%29

Since the three angles of any triangle total 180°, we find the 
remaining angle BAC by adding 50°+20°, getting 70°, then subtracting
from 180° and getting 110°, so we write that in for angle BAC:

drawing%28400%2C400%2C-1.5%2C1.5%2C-1.5%2C1.5%2C+circle%280%2C0%2C1%29%2C%0D%0Atriangle%28-1%2C0%2C-.7660444431%2C.6427876097%2C.17364818%2C-.9848078%29%2C%0D%0Alocate%28-1.1%2C.4%2C%2740%B0%27%29%2C+locate%28-.82%2C.45%2C%2750%B0%27%29%2C+locate%28-1.1%2C.1%2C%27A%27%29%2C%0D%0Alocate%28-.9%2C.7%2C%27B%27%29%2Clocate%28.2%2C-1%2C%27C%27%29%2Clocate%28-.27%2C-.5%2C%2720%B0%27%29%2C%0D%0Alocate%28-.95%2C.1%2C%27110%B0%27%29++%29

The inscribed angle at B is 50°. It subtends arc AC, and since it 
is 1%2F2 of the measure of its inscribed arc, the arc AC must be
2x50° or 100°, so we write in 100° for arc AC:

drawing%28400%2C400%2C-1.5%2C1.5%2C-1.5%2C1.5%2C+circle%280%2C0%2C1%29%2C%0D%0Atriangle%28-1%2C0%2C-.7660444431%2C.6427876097%2C.17364818%2C-.9848078%29%2C%0D%0Alocate%28-1.1%2C.4%2C%2740%B0%27%29%2C+locate%28-.82%2C.45%2C%2750%B0%27%29%2C+locate%28-1.1%2C.1%2C%27A%27%29%2C%0D%0Alocate%28-.9%2C.7%2C%27B%27%29%2Clocate%28.2%2C-1%2C%27C%27%29%2Clocate%28-.27%2C-.5%2C%2720%B0%27%29%2C%0D%0Alocate%28-.95%2C.1%2C%27110%B0%27%29%2C+locate%28-.9%2C-.8%2C%27100%B0%27%29++%29

The big major arc going clockwise from B around to C is subtended by 
the 110° angle at A.  And since it is 1%2F2 of the measure of its 
inscribed arc, the large major arc BC must be 2x110° or 220°, so we 
write in 220° for major arc BC, going clockwise from B around to C:

drawing%28400%2C400%2C-1.5%2C1.5%2C-1.5%2C1.5%2C+circle%280%2C0%2C1%29%2C%0D%0Atriangle%28-1%2C0%2C-.7660444431%2C.6427876097%2C.17364818%2C-.9848078%29%2C%0D%0Alocate%28-1.1%2C.4%2C%2740%B0%27%29%2C+locate%28-.82%2C.45%2C%2750%B0%27%29%2C+locate%28-1.1%2C.1%2C%27A%27%29%2C%0D%0Alocate%28-.9%2C.7%2C%27B%27%29%2Clocate%28.2%2C-1%2C%27C%27%29%2Clocate%28-.27%2C-.5%2C%2720%B0%27%29%2C%0D%0Alocate%28-.95%2C.1%2C%27110%B0%27%29%2C+locate%28-.9%2C-.8%2C%27100%B0%27%29%2C+locate%28.9%2C.5%2C%27220%B0%27%29%29

Notice as a partial check that the three arcs have sum 360°.

minor are AB =  40²
minor arc AC = 100°
major arc BC = 220°
-------------------
       total = 360°

Edwin