PreCalculus Honors Unit Test #6

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Part 1: Calculator Allowed

 1 Solve the triangle, given $\alpha =105°$, $a=23$ and $c=14$. (8)

The dreaded angle-side-side version of the Law of Sines. Happily, since the angle is obtuse, we need only check that the side opposite the given angle is longer than the other given side—which is the case ($23>14$). Starting the solution: $\frac{sin105°}{23}=\frac{sin\gamma }{14}⇒\gamma ={sin}^{-1}\left(\frac{14sin105°}{23}\right)\approx 36.012°$. That means $105°+\beta +36.012°=180°⇒\beta =38.988°$. Finally, $\frac{sin105°}{23}=\frac{sin38.988°}{b}⇒b=\frac{23sin38.988°}{sin105°}\approx 14.9811$.

 2 Solve the triangle, given $\gamma =101°$, $a=24$ and $b=29$. (7)

Since this is a case of side-angle-side, the Law of Cosines will work without any issues. ${c}^{2}={24}^{2}+{29}^{2}-\left(2\right)\left(24\right)\left(29\right)cos101°⇒c\approx 41.0196$. ${24}^{2}={29}^{2}+{41.0196}^{2}-\left(2\right)\left(29\right)\left(41.0196\right)cos\left(\alpha \right)$ $⇒\alpha ={cos}^{-1}\left(\frac{{24}^{2}-{29}^{2}-{41.0196}^{2}}{\left(-2\right)\left(29\right)\left(41.0196\right)}\right)\approx 35.0532°$. Thus, $35.0532+\beta +101°=180°⇒\beta \approx 43.9468°$.

 3 Find the area of the triangle pictured above. (5)

Easy! $\left(\frac{1}{2}\right)\left(11\right)\left(14\right)sin128°\approx 60.6768$ square feet.

 PreCalculus Honors Unit Test #6

Part 2: No Calculator Allowed

 4 Sketch the graph of $y=sin\left(2x\right)-1$. (5)

Divide the x-values by 2 and shift down 1.

 5 Sketch the graph of $y=3cos\left(x+\frac{\pi }{4}\right)$. (5)

 6 Sketch the graph of $y=tan\left(2x-\frac{\pi }{3}\right)$. (5)

Divide the x-values by 2 and shift right $\frac{\pi }{6}$.

 7 Rewrite as an algebraic expression: $cot\left({cos}^{-1}\left(x\right)\right)$. (6)

Draw a triangle! The side adjacent will have length x and the hypotenuse will have length 1. Thus, the side opposite is ${x}^{2}+{s}^{2}=1⇒{s}^{2}=1-{x}^{2}⇒s=\sqrt{1-{x}^{2}}$. Cotangent is adjacent over opposite, so the expression becomes $\frac{x}{\sqrt{1-{x}^{2}}}$.

 8 Find the area of a triangle with side lengths of 4, 11 and 12. (5)

Time for Heron’s formula! $s=\left(\frac{1}{2}\right)\left(4+11+12\right)=\frac{27}{2}$, so the area of the triangle is $\sqrt{\left(\frac{27}{2}\right)\left(\frac{27}{2}-4\right)\left(\frac{27}{2}-11\right)\left(\frac{27}{2}-12\right)}$. That’s a really nasty expression, so I think I’ll leave it alone rather than trying to simplify it.

 9 Evaluate: ${tan}^{-1}\left(-1\right)$ (2)

$-\frac{\pi }{4}$

 10 Evaluate: ${cos}^{-1}\left(-\frac{1}{2}\right)$ (2)

$\frac{2\pi }{3}$

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