SAT II Math I : Finding Sides with Trigonometry

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #1 : Trigonometry

A plane flies \displaystyle 20 degrees north of east for \displaystyle 100 miles.  It then turns and flies \displaystyle 20 degrees south of east for \displaystyle 200 miles.  Approximately how many miles is the plane from its starting point?  (Ignore the curvature of the Earth.)

Possible Answers:

\displaystyle 121

\displaystyle 139

\displaystyle 205

\displaystyle 172

\displaystyle 297

Correct answer:

\displaystyle 139

Explanation:

The plane flies two sides of a triangle.  The angle formed between the two sides is 40 degrees.  In a Side-Angle-Side situation, it is appropriate to employ the use of the Law of Cosines.

\displaystyle c^2 = a^2 + b^2 - 2ab \cos C

\displaystyle c^2 = 100^2 + 200^2 - 2(100)(200) \cos 40^{\circ}

\displaystyle c^2 \approx 19358

\displaystyle c \approx \sqrt{19358} \approx 139

Example Question #1 : Trigonometry

In \displaystyle \bigtriangleup ABC:

\displaystyle AB = 13

\displaystyle AC = 17

\displaystyle BC = 25

Evaluate \displaystyle m \angle A to the nearest degree.

Possible Answers:

Insufficient information is provided to answer the question.

\displaystyle 151^{\circ }

\displaystyle 68^{\circ }

\displaystyle 112 ^{\circ }

\displaystyle 141^{\circ }

Correct answer:

\displaystyle 112 ^{\circ }

Explanation:

The figure referenced is below: 

Triangle z

By the Law of Cosines, the relationship of the measure of an angle \displaystyle \gamma of a triangle and the three side lengths \displaystyle a\displaystyle b, and \displaystyle c\displaystyle c the sidelength opposite the aforementioned angle, is as follows:

\displaystyle c^{2} = a ^{2} + b^{2} - 2ab \cos \gamma

All three sidelengths are known, so we are solving for \displaystyle \gamma. Setting

\displaystyle c= BC = 25. the length of the side opposite the unknown angle;

\displaystyle a = AB = 13;

\displaystyle b = AC = 17;

and \displaystyle \gamma = \cos A,

We get the equation

\displaystyle 25^{2} =13^{2} +17^{2} - 2 (13) (17)\cos A

\displaystyle 625 =169 +289- 442 \cos A

\displaystyle 625 =458- 442 \cos A

Solving for \displaystyle \cos A:

\displaystyle 625 - 458 =458- 442 \cos A - 458

\displaystyle 167 = - 442 \cos A

\displaystyle \frac{167}{-442} = \frac{- 442 \cos A}{-442}

\displaystyle \cos A \approx -0.3778

Taking the inverse cosine:

\displaystyle A = \cos ^{-1} -0.3778 \approx 112 ^{\circ },

the correct response.

Example Question #1 : Finding Sides With Trigonometry

In \displaystyle \bigtriangleup ABC:

\displaystyle AB = 19

\displaystyle AC = 25

\displaystyle m \angle A = 57^{\circ }

Evaluate the length of \displaystyle \overline{BC} to the nearest tenth of a unit.

 

Possible Answers:

\displaystyle 38.8

\displaystyle 13.7

\displaystyle 27.0

\displaystyle 24.2

\displaystyle 21.6

Correct answer:

\displaystyle 21.6

Explanation:

The figure referenced is below: 

Triangle z

By the Law of Cosines, given the lengths \displaystyle a and \displaystyle b of two sides of a triangle, and the measure \displaystyle \gamma of their included angle, the length \displaystyle c of the third side can be calculated using the formula

\displaystyle c^{2} = a ^{2} + b^{2} - 2ab \cos \gamma

Substituting \displaystyle a = AB = 19\displaystyle b = AC = 25, \displaystyle c = BC, and \displaystyle \gamma = m \angle A = 57^{\circ }, then evaluating:

\displaystyle (BC)^{2} = 19 ^{2} + 25 ^{2} - 2 (19)(25)\cos 57^{\circ }

\displaystyle (BC)^{2} \approx 361 + 625 - 950 \cdot 0.5446

\displaystyle (BC)^{2} \approx 361 + 625 - 517.4

\displaystyle (BC)^{2} \approx 468.6

Taking the square root of both sides:

\displaystyle BC \approx \sqrt{468.6} \approx 21.6.

Example Question #3 : Finding Sides With Trigonometry

In \displaystyle \bigtriangleup ABC:

\displaystyle m \angle A = 76 ^{\circ }

\displaystyle m \angle B = 65 ^{\circ }

\displaystyle AB = 34

 \displaystyle AC 

Possible Answers:

\displaystyle 49

\displaystyle 24

\displaystyle 63

\displaystyle 18

\displaystyle 52

Correct answer:

\displaystyle 49

Explanation:

The figure referenced is below:

Triangle z

The Law of Sines states that given two angles of a triangle with measures \displaystyle \alpha, \beta, and their opposite sides of lengths \displaystyle a,b, respectively,

\displaystyle \frac{\sin \beta}{b} = \frac{\sin \gamma}{c},

or, equivalently,

\displaystyle \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}.

In this formula, we set:

\displaystyle b = AC, the desired sidelength;

\displaystyle \beta =m \angle B = 65^{\circ }, the measure of its opposite angle;

\displaystyle c =AB = 34, the known sidelength;

\displaystyle \gamma =m \angle C, the measure of its opposite angle, which is

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

\displaystyle = 180 ^{\circ } - ( 76^{\circ } + 65 ^{\circ })

\displaystyle = 180 ^{\circ } - 141 ^{\circ }

\displaystyle =39 ^{\circ }

Substituting in the Law of Sines formula and solving for \displaystyle b:

\displaystyle \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}

\displaystyle \frac{b}{\sin 65 ^{\circ }} = \frac{ 34}{\sin 39 ^{\circ }}

\displaystyle \frac{b}{\sin 65 ^{\circ }} \cdot \sin 65 ^{\circ } = \frac{ 34}{\sin 39 ^{\circ }} \cdot \sin 65 ^{\circ }

\displaystyle b= \frac{ 34}{\sin 39 ^{\circ }} \cdot \sin 65 ^{\circ }

Evaluating the sines, then calculating:

\displaystyle b\approx \frac{ 34}{0.6293 } \cdot 0.9063

\displaystyle b \approx 49

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