First Question
For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.
We dropped perpendiculars from C and D to intersect AB at Q and P respectively.
As can be seen in [tex] \Delta BCQ [/tex], we can easily find the values of CQ and BQ.
Since, [tex] Sin(75^0)=\frac{CQ}{8} [/tex]
[tex] \therefore CQ=8\times Sin(75^0)\approx 7.73 [/tex] ft
In a similar manner we can find BQ as:
[tex] Cos(75^0)=\frac{BQ}{8} [/tex]
[tex] BQ\approx2.07 [/tex] ft
All these values can be found in the diagram attached.
Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:
[tex] PQ=22-(AP+QB)=22-(2.07+2.07)=17.86 [/tex]
Let us now consider[tex] \Delta AQC [/tex]
We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of [tex] \Delta AQC [/tex].
[tex] AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38 [/tex] feet.
Thus, out of the given options, Option B is the closest and hence is the answer.
Second Question
For this question we can directly apply the formula for the area of a triangle using sines which is as:
Area=[tex] \frac{1}{2} [/tex](First Side)(Second Side)(Sine of the angle between the two sides)
Thus, from the given data,
Area=[tex] \frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845 [/tex] [tex] m^2 [/tex]
Therefore, Option D is the correct option.
Third Question
For this question we will apply the Sine Rule to the [tex] \Delta ABC [/tex] given to us.
Thus, from the triangle we will have:
[tex] \frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)} [/tex]
[tex] \frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)} [/tex]
[tex] \frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)} [/tex]
This gives [tex] a [/tex] to be:
[tex] a\approx28.44 [/tex]
Which is not close to any of the given options.
Fourth Question
Please find the second attachment for a better understanding of the solution provided her.
As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.
[tex] AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)} [/tex]
[tex] \therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8 [/tex] miles.
Thus, Option D is the answer.
