A spider ate 25\%25%25, percent more bugs this month than last month. The spider ate 8 bugs last month. How many bugs did the sp
1 answer:
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Answer:
Use the angle copy procedure to copy the angles to the ends of c.
Step-by-step explanation:
An angle is copied with a straightedge two settings of a compass.
- Set the compass to an arbitrary radius. An appropriate choice is a radius that is half or more of the length of the shortest ray of the angles you want to copy.
- Put the point of the compass at the vertex of an angle you want to copy. Using that same radius, draw arcs through both rays of the angle. Do this for all the angles you want to copy.
- Put the point of the compass at the place where you want the vertex of the copied angle. Here, that is either (both) end points of segment c. (You might want to label the ends of segment c as "A" and "B" so you know which angle you're copying where.) Using the same radius as before, draw an arc through the segment and through the space where you expect the ray from the copied angle to lie.
- For one of the source angles, set the compass radius to the distance between the points where the first arc crosses the angle's rays. Then, put the point of the compass at the place on the segment c where the corresponding arc crosses. Use the compass to mark a point on that arc the same distance as on the source angle. Draw a line from the vertex through the point you just marked. That line will make the same angle with c as the original angle.
- Repeat step 4 for the other angle you want to copy, at the other end of segment c. In general, the compass setting will be different (unless all the angles have the same measure).
The place where the rays from the copied angles cross is the third vertex (vertex C) of the triangle you're constructing.
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<em>Comments on the attached diagram</em>
In the attached diagram, "step 1" is to place the target vertex. You already have that as one end of segment C. The arcs numbered 2 and 3 in the diagram are the arcs resulting from executing steps 2 and 3 above. (They have arbitrary radius "r", which is the same everywhere.) You will have two sets, because you are copying two angles.
The arcs numbered 4 and 5 in the diagram have radius ST, the distance you set in step 4 above. That distance is used as the radius of arc 5, so the length VW will be the same as the length ST. The straightedge is used to draw a line through B and W, completing the copy of the angle.
Answer:
m∠JKP = 31.5°
Step-by-step explanation:
Incenter of a triangle is the point where all the bisectors of interior angles intersect each other.
JN is the angle bisector of ∠KJL.
Therefore, m∠KJN = m∠LJN
(7x - 6) = (5x + 4)
7x - 5x = 6 + 4
2x = 10
x = 5
m∠KJN = (7x - 6)
= 7(5) - 6
= 35 - 6
= 29°
In ΔKJN,
m∠JKN + m∠KNJ + m∠NJK = 180°
m∠JKN + 90° + 29° = 180°
m∠JKN = 180°- 119° = 61°
Since KO is the angle bisector of ∠JKN,
m∠JKP =
=
= 30.5°
