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4 pages/≈1100 words

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Level:

MLA

Subject:

Literature & Language

Type:

Math Problem

Language:

English (U.S.)

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# MATHEMATICS SOLUTIONS (Math Problem Sample)

Instructions:

THESE MATHEMATICS SOLUTIONS COMPRISES THE FORMULAR TO SOLVE THE AREA OF A TRIANGLE BY THE USE OF TRIGONOMTRY METHOD AND THE WHOLE SOLUTIONS WAS WRITTEN USING MODERN LANGUAGE ASSOCIATION(MLA) style.
the final answers was boldly written with underline.
unlike the conventional methods of solving area of a triangle, sine,cosine ,and tangent of an angle can equally be used to provide solution as long as an angle is given in degree from,those are the example displayed in the presented solutions. source..

Content:

Student’s Name
Professor’s Name
Subject
Date
Area of Triangle
1
Tan 48o = 1.1
Height = 1.1 x 41 = 45.1
Tan 62o = 1.9
Base of the right triangle = 1.9 x 62 = 117.8
Base of the triangles = 117.8 + 41 = 158.8
Area = 0.5 x 158.8 x 45.1 = 3,580.9
2
Tan 32o = 0.6
Height = 28/0.6 = 46.7
Tan 26o = 0.5
Base of the left triangle = 46.7 / 0.5 = 93.4
Base of the triangles = 93.4 + 28 = 121.4
Area = 0.5 x 121.4 x 46.7 = 2834.7
3
Tan 55o = 1.4
Height = 1.4 x 9 = 12.6
Tan 42o = 0.9
Base of the right triangle = 12.6 / 0.9 = 14
Base of the triangles = 14 + 9 = 23
Area = 0.5 x 23 x 12.6 = 144.9
4
Tan 36o = 0.7
Height = 36 / 0.7 = 51.4
Tan 64o = 2.1
Base of the right triangle = 51.4 / 2.1 = 24.5
Base of the triangles = 36 + 24.5 = 60.5
Area = 0.5 x 60.5 x 51.4 = 1,554.9
5
Tan 21o = 0.4
Height = 0.4 x 48 = 19.2
Tan 46o = 1
Base of the left triangle = 19.2 x 1 = 19.2
Base of the triangles = 48 + 19.2 = 67.2
Area = 0.5 x 67.2 x 19.2 = 645.1
6
Tan 64o = 2.1
Height = 5 / 2.1 = 2.4
Tan 70o = 2.7
Base of the left triangle = 2.4 / 2.7 = 0.9
Base of the triangles = 5 + 0.9 = 5.9
Area = 0.5 x 5.9 x 2.4 = 7.1
7
Tan 38o = 0.8
Height = 23 / 0.8 = 28.8
Tan 26o = 0.5
Base of the left triangle = 28.8 / 0.5 = 57.6
Base of the triangles = 57.6 + 23 = 80.6
Area = 0.5 x 80.6 x 28.8 = 1,160.6
8
Tan 48o =...

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