When a transversal such as \(\displaystyle t\) crosses two parallel lines, two corresponding angles - angles in the same relative position to their respective lines - are congruent. Therefore, 

\(\displaystyle 3x+2y = 4x+21\)

\(\displaystyle 3x+2y- 3x - 21 = 4x+21 - 3x - 21\)

\(\displaystyle x = 2y - 21\)

Two same-side interior angles are supplementary - that is, their angle measures total 180 - so

\(\displaystyle 3x+2y + 2y-27 = 180\)

\(\displaystyle 3x+4y-27 = 180\)

\(\displaystyle 3x+4y= 207\)

We can solve this system by the substitution method as follows:

\(\displaystyle 3( 2y - 21)+4y= 207\)

\(\displaystyle 6y-63+4y= 207\)

\(\displaystyle 10y-63= 207\)

\(\displaystyle 10y = 270\)

\(\displaystyle y = 27\)

Backsolve:

\(\displaystyle x = 2y - 21\)

\(\displaystyle x = 2 (27)- 21 = 54-21 = 33\)

\(\displaystyle x+y = 27+33 = 60\), which is the correct response.

The top and bottom angles, being vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure, so 

\(\displaystyle 2x+y = 2y\), 

or, simplified,

\(\displaystyle 2x+y - y= 2y - y\)

\(\displaystyle y = 2x\)

The right and bottom angles form a linear pair, so their degree measures total 180. That is, 

\(\displaystyle 2y+ 8x = 180\)

Substitute \(\displaystyle 2x\) for \(\displaystyle y\):

\(\displaystyle 2(2x)+ 8x = 180\)

\(\displaystyle 4x+8x = 180\)

\(\displaystyle 12x= 180\)

\(\displaystyle x = 15\)

The left and right angles, being vertical angles, have the same measure, so, since the right angle measures \(\displaystyle \left (8x \right )^{\circ } =\left (8 \cdot 15 \right )^{\circ } = 120 ^{\circ }\), this is also the measure of the left angle, \(\displaystyle \angle 1\).

By the Same-Side Interior Angle Theorem, since \(\displaystyle \overline{TR} || \overline{AP}\), \(\displaystyle \angle T\) and \(\displaystyle \angle P\) are supplementary - that is, their degree measures total \(\displaystyle 180 ^{\circ }\). Therefore, 

\(\displaystyle m \angle P + m \angle T = 180 ^{\circ }\)

\(\displaystyle m \angle P = 180 ^{\circ } - m \angle T\)

\(\displaystyle m \angle P = 180 ^{\circ } - 106 ^{\circ } = 74^{\circ }\)

\(\displaystyle \angle P\) is an inscribed angle, so the arc it intercepts, \(\displaystyle \overarc{TA}\), has twice its degree measure;

\(\displaystyle m \overarc{TA} = 2 \cdot m \angle P =2 \cdot 74^{\circ } = 148 ^{\circ }\).

The corresponding major arc, \(\displaystyle \overarc{TPA}\), has as its measure 

\(\displaystyle m \overarc{TPA} = 360 ^{\circ }- m \overarc{TA} = 360 ^{\circ }- 148 ^{\circ }= 212 ^{\circ }\)

The measure of an angle formed by two tangents to a circle is equal to half the difference of those of its intercepted arcs:

\(\displaystyle m \angle R = \frac{1}{2} \left (m \overarc{TPA} - m \overarc{TA} \right )\)

\(\displaystyle m \angle R = \frac{1}{2} \left (212 ^{\circ } -148 ^{\circ } \right )= \frac{1}{2} \cdot 64 ^{\circ } = 32^{\circ }\)

Again, by the Same-Side Interior Angles Theorem, \(\displaystyle \angle R\) and \(\displaystyle \angle A\) are supplementary, so

\(\displaystyle m \angle A + m \angle R= 180 ^{\circ }\)

\(\displaystyle m \angle A = 180 ^{\circ } - m \angle R\)

\(\displaystyle m \angle A = 180 ^{\circ } - 32 ^{\circ } = 148^{\circ }\)

An isosceles triangle, by definition, has two sides of equal length. Having the third side measure either 3 feet or 4 feet would make the triangle meet this criterion.

3 feet is equal to \(\displaystyle 3 \times 12 = 36\) inches, and 4 feet is equal to \(\displaystyle 4 \times 12 = 48\) inches. We choose 36 inches, since that, but not 48 inches, is a choice.

Because the triangles are similar, proportions can be used to solve for the length of the side:

\(\displaystyle \frac{9}{3}=\frac{6}{x}\)

Cross-multiply:

\(\displaystyle 9x=18\)

\(\displaystyle x=2\)

The base angles of an isosceles triangle are always equal. Therefore both base angles are \(\displaystyle 42^{\circ}\).

Let \(\displaystyle x=\) the measure of the third angle. Since the sum of the angles of a triangle is \(\displaystyle 180^{\circ}\), we can solve accordingly:

\(\displaystyle 42+42+x=180\Rightarrow x=180-84\Rightarrow x=96^{\circ}\)

To find the missing side, use the Pythagorean Theorem \(\displaystyle (a^2+b^2=c^2)\). Plug in (remember c is always the hypotenuse!) so that \(\displaystyle 6^2+b^2=10^2\). Simplify and you get \(\displaystyle 36+b^2=100.\) Subtract 36 from both sides so that you get \(\displaystyle b^2=64.\)Take the square root of both sides. B is 8.

By the Pythagorean Theorem,

\(\displaystyle x^{2}+ (x+3)^{2} = 20^{2}\)

\(\displaystyle x^{2}+ x^{2}+6x+9= 400\)

\(\displaystyle 2x^{2}+6x-391 = 0\)

By the Pythagorean Theorem,

\(\displaystyle x^{2}+ (x+5)^{2} = (x+7)^{2}\)

\(\displaystyle x^{2}+ x^{2} +10x+25 = x^{2} +14x +49\)

\(\displaystyle 2x^{2} +10x+25 = x^{2} +14x +49\)

\(\displaystyle x^{2} -4x-24= 0\)

First, find \(\displaystyle AB\).

Since \(\displaystyle \overline{DF }\) is an altitude of right \(\displaystyle \Delta ADB\) to its hypotenuse, 

\(\displaystyle \frac{FB}{FD}= \frac{FD}{AF}\)

\(\displaystyle \frac{FB}{8}= \frac{8}{6}\)

\(\displaystyle FB= \frac{8}{6} \cdot 8 = 10 \frac{2}{3}\)

\(\displaystyle AB = AF + FB = 6 + 10 \frac{2}{3} = 16 \frac{2}{3}\)

\(\displaystyle \Delta AFD \sim \Delta ABC\) by the Angle-Angle Postulate, so 

\(\displaystyle \frac{BC}{FD} = \frac{AB}{AF}\)

\(\displaystyle \frac{BC}{8} = \frac{16 \frac{2}{3}}{6}\)

\(\displaystyle BC = 16 \frac{2}{3}\cdot 8 \div 6 = 22\frac{2}{9}\)