# please solve problem and show work for comparison. Thank you.

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There is not enough information to get a value for a. We can get most of the other angles in terms of a:

b=110-a either from the angle sum of a triangle, or since the angle adjacent to a is an alternate interior angle to b, we can use the fact that a+b is supplementary to the given 70 degree angle since same-side interior angles are supplementary.

c=a+40 since a+b=110 and b+c=150 (same side interior angles are supplementary so c+b+30=180) or the angle sum of a triangle is 180 so b+c+30=180 and substitute 110-a for b.

e=70-a since the angle sum of a triangle is 180 -- using vertical angles we have a triangle with vertices of measure e,c, and 70; substitute for c.

d=110+a since d is an exterior angle of the triangle used for e -- the measure of an exterior angle of a triangle is equal to the sum of the remote interior angles.

f=(110-a)/2 assuming the angles marked f are congruent. 2f is an exterior angle of a triangle with remote interior angles of measure 70 and c; or 2f+e=180 as they form a linear pair and substitute for e.

g=107 assuming that the diagram indicates that both angles marked s are congruent. Use the "shepherds crook" problem procedure -- draw a line through the angle measured 106 degrees parallel to the line 3. Then use alternate interior angles, or same-side interior angles to get the measure of the angles marked s.

h=125+a/2 since we have a quadrilateral with angles of h,f,110, and 70. Substitute for f.

i=24. 2x+r=180 and 3x+i=180 so r+i+5x=360. However, the triangle with vertices at r and i has the third angle of measure 80 so r+i=100. Thus x=52 and i=24.

j=110 We are given a 30 degree angle making n=150 degrees. Then using same-side interior angles we have a 30 degree angle in the small triangle with j as a vertex. The given angle of 100 degrees gives a second angle of this triangle as 80 degrees. j is the exterior angle of this triangle.

k=110 corresponds with j.

m=a+40 as m=c by alternate interior angles.

n=150 as it forms a linear pair with the given 30 degree angle.

p=110 The 80 degree angle of the small triangle with j has a corresponding angle, then vertical angles. The given 30 degree angle has a vertical angle in the same triangle. p is an exterior angle of this triangle.

q=80 vertical to the 80 degree angle of the little triangle with j as a vertex.

r=76 we computed x earlier.

s=107 assuming that the angles marked as s and sg are congruent.

If we assume a value for a we can get numerical values for all of the angles. The obvious assumption is a=30, but this implies that the transversal forming a is parallel to lines 1 and 2 and this is not given nor can it be shown.

Also, there is some confusion in the diagram as to angle g as well as angle f.