To find the other angle, subtract the given angle from \(\displaystyle 90$^{\circ}$\) since complementary angles add up to \(\displaystyle 90$^{\circ}$\).

The complementary is:

\(\displaystyle \angle 90-\angle65 = \angle25\)

Supplementary angles add up to \(\displaystyle 180$^{\circ}$\). In order to find the correct angle, take the known angle and subtract that from \(\displaystyle 180$^{\circ}$\).

\(\displaystyle 180$^{\circ}$ -100$^{\circ}$ = 80$^{\circ}$\)

The complement to an angle is ninety degrees subtract the angle since two angles must add up to 90.  In this case, since we are given the angle in radians, we are subtracting from \(\displaystyle \frac{\pi}{2}\) instead to find the complement.  The conversion between radians and degrees is:  \(\displaystyle \pi \textup{ radians } = 180 \textup{ degrees}\)

\(\displaystyle \frac{\pi}{2}-\frac{\pi}{6}\)

Reconvert the fractions to the least common denominator.

\(\displaystyle \frac{\pi}{2}-\frac{\pi}{6} = \frac{3\pi}{6}-\frac{\pi}{6} = \frac{2\pi}{6}\)

Reduce the fraction.

\(\displaystyle \frac{2\pi}{6} = \frac{\pi}{3}\)

The size of the regions does not matter here. What matters is the angle measurement, or, equivalently, what part of a circle each sector is.

The two smaller red regions each comprise one fourth of one fourth of a circle, or 

\(\displaystyle \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\) circle.

The two larger red regions each comprise one third of one fourth of a circle, or

\(\displaystyle \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\) circle.

Therefore, the total angle measure comprises

\(\displaystyle \frac{1}{16} + \frac{1}{16} + \frac{1}{12} + \frac{1}{12}\)

\(\displaystyle =\frac{3}{48} + \frac{3}{48} + \frac{4}{48} + \frac{4}{48}\)

\(\displaystyle = \frac{14}{48} = \frac{7}{24}\)

of a circle.

This makes \(\displaystyle \frac{7}{24}\) the correct probability.

The size of the regions does not matter here. What matters is the angle measurement, or, equivalently, what part of a circle each sector is.

Two of the blue sectors are each one third of one quarter-circle, and thus are

\(\displaystyle \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}\)

of one circle. 

The other two blue sectors are each one fourth of one quarter-circle, and thus are

\(\displaystyle \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\)

of one circle. 

Therefore, the total angle measure comprises

\(\displaystyle \frac{1}{16} + \frac{1}{16} + \frac{1}{12} + \frac{1}{12}\)

\(\displaystyle =\frac{3}{48} + \frac{3}{48} + \frac{4}{48} + \frac{4}{48}\)

\(\displaystyle = \frac{14}{48} = \frac{7}{24}\)

of a circle. This makes \(\displaystyle \frac{7}{24}\) the correct probability. As odds, this translates to 

\(\displaystyle (24 -7): 7\), or \(\displaystyle 17:7\) odds against the spinner landing in blue.

The size of the regions does not matter here. What matters is the angle measurement, or, equivalently, what part of a circle each sector is.

The purple region is one third of one quarter of a circle, or, equvalently,

\(\displaystyle \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}\) 

of a circle, so its central angle is \(\displaystyle \frac{1}{12}\) of the total measures of the angles of the sectors. This makes \(\displaystyle \frac{1}{12}\) the probability of the spinner stopping inside the purple region; this translates to

\(\displaystyle (12-1):1\) or \(\displaystyle 11:1\) odds against this occurrence.

\(\displaystyle \angle A\) and \(\displaystyle \angle B\) are a pair of adjacent angles of the parallelogram, and as such, they are supplementary - that is, their degree measures total 180. Therefore, 

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

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

\(\displaystyle = 180 ^{\circ } - (t+ 56 )^{\circ }\)

\(\displaystyle = (180 - 56 - t )^{\circ }\)

\(\displaystyle = (124-t )^{\circ }\)

The sum of the measures of the angles of a triangle is 180 degrees, so solve for \(\displaystyle t\) in the equation:

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

\(\displaystyle (3t-14) + (2t+16) + 5t = 180\)

\(\displaystyle 10t+ 2 = 180\)

\(\displaystyle 10t+ 2- 2 = 180 - 2\)

\(\displaystyle 10t = 178\)

\(\displaystyle 10t \div 10 = 178 \div 10\)

\(\displaystyle t = 17.8\)

\(\displaystyle m \angle A = (3t-14)^{\circ } = (3 \cdot 17.8-14)^{\circ } = (53.4-14)^{\circ } = 39.4^{\circ }\)

\(\displaystyle m \angle B = (2t+16)^{\circ } = (2 \cdot 17.8 + 16)^{\circ }= (35.6+ 16) ^{\circ }= 51.6^{\circ }\)

\(\displaystyle m \angle C =( 5t)^{\circ } = ( 5 \cdot 17.8)^{\circ } = 89 ^{\circ }\)

All three angles measure less than 90 degrees and are therefore acute angles; that makes \(\displaystyle \bigtriangleup ABC\) an acute triangle.

The total degrees of the angles in a triangle are \(\displaystyle 180\).  

Since it is a right triangle, one of the three angles must be \(\displaystyle 90.\) 

That leaves you with \(\displaystyle 90\) for the other two angles \(\displaystyle (180-90=90)\).  

If they are equal, you just divide the remaining degrees by \(\displaystyle 2\) to get \(\displaystyle 90/2=45\).

A right triangle has one 90 degree angle and all three angles must equal 180 degrees.

To find the answer, just subtract the two angles you have from the total to get 

\(\displaystyle 180^\circ-\textup{Angle 1}-\textup{Angle 2}=\textup{Angle 3}\)

The angle we have are,

\(\displaystyle \\ \textup{Angle 1}=90^\circ \\ \textup{Angle 2}=45^\circ\).

Substituting these into the formula results in the solution.

\(\displaystyle 180^\circ-90^\circ-45^\circ=45^\circ\).