I'll do problems 13 and 14 to get you started.
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Problem 13
<h3>Answer: 32 degrees</h3>
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Explanation:
Refer to the diagram below for problem 13. Since BD is a tangent of the circle, this means angle ABD is 90 degrees, and furthermore x+y = 90. This solves to y = 90-x.
If we focus on triangle ABC, then we can add the interior angles and set the sum equal to 180
A+B+C = 180
z+y+y = 180
z+2y = 180
z+2(90-x) = 180 .... plug in y = 90-x
z+180-2x = 180
z-2x = 180-180
z-2x = 0
z = 2x
We'll use this later.
Now move onto triangle ABD. This is a right triangle due to angle ABD being 90 degrees. The acute angles z and 26 are complementary, meaning they add to 90. So,
z+26 = 90
z = 90-26
z = 64
Now plug this into z = 2x and solve for x
z = 2x
64 = 2x
2x = 64
x = 64/2
x = 32
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Problem 14
<h3>Answer: 31 degrees</h3>
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Explanation:
Refer to the image shown below for problem 14. Like before, I've added various labels to it to help find the angle we're after (the red angle x).
Note how I've drawn a dashed line from point A to point D. This splits up quadrilateral ABCD into two triangles ABD and ACD. These triangles can be shown to be congruent using the HL (hypotenuse leg) theorem.
The congruent triangles also implies that angle DBC and angle BCD are the same measure (both are shown in green as the variable y).
Focus on triangle BCD. It has interior angles of: y, y and 62. Add them up, set the sum equal to 180.
y+y+62 = 180
2y+62 = 180
2y = 180-62
2y = 118
y = 118/2
y = 59
As the image attachment mentions, the angles x and y add to 90 degrees. This is because angle ABD is 90 degrees, due to segment BD being tangent to the circle. So x+y = 90 leads to x = 90-y, and therefore,
x = 90-y
x = 90-59
x = 31
Side note: it is not a coincidence that we ended up with half of the value of 62.
