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SAT II Math I : Finding Sides with Trigonometry

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Example Questions

Example Question #1 : Trigonometry

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

Possible Answers:

297

205

139

121

172

Correct answer:

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.

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

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

$c^2 \approx 19358$

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

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Example Question #1 : Trigonometry

In $\bigtriangleup ABC$ :

AB = 13

AC = 17

BC = 25

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

Possible Answers:

Insufficient information is provided to answer the question.

$151^{\circ }$

$68^{\circ }$

$112 ^{\circ }$

$141^{\circ }$

Correct answer:

$112 ^{\circ }$

Explanation:

The figure referenced is below:

Triangle z

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

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

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

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

a = AB = 13 ;

b = AC = 17 ;

and $\gamma = \cos A$ ,

We get the equation

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

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

$625 =458- 442 \cos A$

Solving for $\cos A$ :

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

$167 = - 442 \cos A$

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

$\cos A \approx -0.3778$

Taking the inverse cosine:

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

the correct response.

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Example Question #1 : Finding Sides With Trigonometry

In $\bigtriangleup ABC$ :

AB = 19

AC = 25

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

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

Possible Answers:

38.8

13.7

27.0

24.2

21.6

Correct answer:

21.6

Explanation:

The figure referenced is below:

Triangle z

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

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

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

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

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

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

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

Taking the square root of both sides:

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

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Example Question #3 : Finding Sides With Trigonometry

In $\bigtriangleup ABC$ :

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

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

AB = 34

$\textup{Evaluate}$ $\textup{to the nearest whole unit.}$

Possible Answers:

Correct answer:

Explanation:

The figure referenced is below:

Triangle z

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

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

or, equivalently,

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

In this formula, we set:

b = AC , the desired sidelength;

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

c =AB = 34 , the known sidelength;

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

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

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

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

$=39 ^{\circ }$

Substituting in the Law of Sines formula and solving for :

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

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

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

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

Evaluating the sines, then calculating:

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

$b \approx 49$

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SAT II Math I : Finding Sides with Trigonometry

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

All SAT II Math I Resources

Example Questions

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Finding Sides With Trigonometry

Example Question #3 : Finding Sides With Trigonometry

All SAT II Math I Resources

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