Trignometry: The Most Common Applications Of Trigonometry

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Applications of Trignometry Trigonometry is the branch of mathematics that is based on the study of triangles. This study helps defining the relations between the different angle measures of a triangle with the lengths of their sides. Trigonometry functions such as sine, cosine, and tangent, and their reciprocals are used to find the unknown parts of a triangle. Laws of sines and cosines are the most common applications of trigonometry that we have used in our pre-calculus class. Historically. Trigonometry was developed for astronomy and geography as it helped early explorers plot the stars and navigate the seas, but scientists have been using it for centuries for other purposes, too. Besides other fields of mathematics, it is used in physics, …show more content…

After the splitting of the quadrilateral. I named the upper triangle as Triangle I and bottom one as Triangle II. First, I found the area of Triangle I using the equation of K=½(280)(320)sin 88° and it gave me approximate answer 44,772.71m2. In order to find the area of Triangle II, I needed to find the length of segment BC and CD; however, I could not do this without the measures of angle DBC and angle BDC . I used the law of sines to find the adjacent angles of DBC. I multiplied the inverse of sine to (sine 88°)(320) and divided it by 417.79 to find the angle of ABD, which is adjacent angle of DBC. I subtracted 49.95° from 102° and got 52.05° as the measure of DBC. Once I got the angle DBC, I could easily find the angle of BDC. I subtracted measure of angle DBC, 32.05°, and the measure of angle C, 90°, from 180° and got 37.95° for the measure of the BDC. By using the law of sines again, I found the lengths of segment CD and BC. I divided (417.79) (sin 52.05°) by sin 90° to get the length of segment CD which was 329.45 m. the length of segment BC was 256.93 m which was also divided by 90° from the product of (417.79) (sin 37.95°). Finally all the information that was to find the area of Triangle II was found, I just had to plug the numbers into the formula, K=12(256.93)(329.45) sin 90°and I got 42,322.79 m2. Last part of this problem was to add the two areas of the triangle to find the area of the equilateral, by adding them together, 44,772

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