ENGTRIG: LECTURE # 4.2 Spherical Trigonometry Spherical Trigonometry Engr. Christian Pangilinan Areas of a Spherical Triangle A= π R2 E 180o E R E = A + B + C − 180o Where: spherical excess radius of the sphere Spherical Triangles Part of the surface of the sphere bounded by three arcs of three great circles Right Spherical Triangle – a spherical triangle containing at least one right angle If the sides are known instead of the angles‚ then L’Huiller’s Formula can be used
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1/20/2015 Honors Geometry Exam Review Chapter 7 flashcards | Quizlet Honors Geometry Exam Review - Chapter Ready to study? Start with Flashcards 7 22 terms by shweta101 Pythagorean Theorem In a right triangle‚ the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. (a^2 + b^2 = c^2 (Page 433) Pythagorean Triple A set of 3 positive integers A‚ B‚ and C that satisfy the equation A^2 + B^2 = C^2 [Ex. (3‚4‚5) (5‚12‚13) (8‚ 15‚17) and (7‚24
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terms of sine and cosine. |tan t = |sin t | |cot t = |1 |= |cos t | | |[pic] | | |[pic] | |[pic] | | |cos t | | |tan t | |sin t | |sec t = |1 | |csc t = |1 | | | | |[pic] | | |[pic] | | | | |cos t | | |sin t | | | The Pythagorean formula for sines and cosines. sin2 t +
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first thing to do to solve the triangle is to use the Law of Sines. ______4. The Law of Sines states that the ratio of the sine of an angle in a triangle to its opposite side is equal to the ratios of the sines of the other two angles to their opposite sides. ______5. Law of Cosines says that the square of any side of a triangle is equal to the squares of the sum of the other two sides‚ minus twice the product of those two sides times the cosine of the included angle. ______6. If you are given the
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We can find OP’‚ we have the tools. The two tools that we will be using mainly are the sine law and cosine law (respectively); Sin Aa = Sin Bb = Sin Cc a2= b2+ c2-2bc cosA and the two triangles that are going to be used; For the first calculations‚ r=1 and OP =2. By finding the ∠O in one triangle‚ I have found the ∠O in both triangles‚ allowing me a complete ration to perform the sine law. Side Note: All Final Answers are rounded to 3 Significant Figures. For the first calculations
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dreadful sin‚ cosine‚ and tangent. This particular rule that also follows through Pre-Calculus could have been used to find the right speed and landing needed to land the plane safely. Using the cosine rule‚ you can find the ground speed using c2=a2+b2=2abcosC . 1002=202+b2-2’20’bcos126. B2-23.5b-9600=0. Once you found the ground speed you could use the sine rule to calculate the heading that the plane should follow: sin=sin=86.7sin126. You can use math with Law of sine and cosine‚ in combinations
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to the nearest hundredth. 1. sin A 3. cos A SOLUTION: The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse. So‚ ANSWER: SOLUTION: The sine of an angle is defined as the ratio of the opposite side to the hypotenuse. So‚ 4. tan A ANSWER: 2. tan C SOLUTION: The tangent of an angle is defined as the ratio of the opposite side to the adjacent side. So‚ ANSWER: 3. cos A SOLUTION: The cosine of an angle is defined as the ratio of the adjacent side to
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First I will state the Cosine Law which deals with finding the side of a triangle if the sides and angles are given. Consider the following figure- In the above given triangle ABC‚ sides AB‚ BC‚ and CA are represented by the small letters c‚ a and b. These are used as it is a
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trigonometry and trigonometric functions. For instance‚ the technique of triangulation is used in astronomy to measure the distance to nearby stars‚ in geography to measure distances between landmarks‚ and in satellite navigation systems. The sine and cosine functions are fundamental to the theory of periodic functions such as those that describe sound and light waves. Fields which make use of trigonometry or trigonometric functions include astronomy (especially‚ for locating the apparent positions of
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I looked at the triangle that is already drawn in the above figure‚ ΔAOP. We know that this triangle is also isosceles because OP=AP. By that logic‚ ∠A=∠O. Using the law of cosines c^2=a^2+b^2-2abcos(C)‚ which works for any triangle‚ I assigned θ to ∠O and determined that cos(θ)=1/(2*OP). Then‚ using the law of sines (insert law of sines here)‚ sin(θ)/1=sin(180-2θ)/OP’ OP’=sin(180-2θ)/sin(θ) OP’=sin(2θ)/sin(θ) OP’=2cos(θ) But because cos(θ)=1/2OP as earlier discovered; OP’=1/OP By using
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