Trigonometry 6 - Ambiguous Case_1
... If there are two possible triangles, then you can use the Sine Law to solve for both of them. The angle opposite the longer side will have two possibilities: either acute or obtuse. They will have the same ___________________ angle. e.g. 5) In FGH , f 6.5 m, g 5.1 m, and G 41 . Draw a diagr ...
... If there are two possible triangles, then you can use the Sine Law to solve for both of them. The angle opposite the longer side will have two possibilities: either acute or obtuse. They will have the same ___________________ angle. e.g. 5) In FGH , f 6.5 m, g 5.1 m, and G 41 . Draw a diagr ...
Chapter 5 notes Trigonometric Ratios Inverse Trigonometry
... This property is called the Law of Sines. This is a powerful tool because you can use the sine ratio to solve for measures of angles and lengths of sides of any triangle, not just right triangles. The law works for angle measures between 0° and 180°. ...
... This property is called the Law of Sines. This is a powerful tool because you can use the sine ratio to solve for measures of angles and lengths of sides of any triangle, not just right triangles. The law works for angle measures between 0° and 180°. ...
Section 4.2
... Congruent Figures: have the same shape, parts and measures. Corresponding parts : are parts of a figure that are related to the same parts of another figure A ...
... Congruent Figures: have the same shape, parts and measures. Corresponding parts : are parts of a figure that are related to the same parts of another figure A ...
5_5_Inequalities_of_One_Triangle
... Use the Ruler Postulate to locate a point D on AC such that DA = BA. Then draw the segment BD. In the isosceles triangle ∆ABD, 1 ≅ 2. Because mABC = m1+m3, it follows that mABC > m1. Substituting m2 for m1 produces mABC > m2. Because m2 = m3 + mC, m2 > mC. Finally because mABC > m2 ...
... Use the Ruler Postulate to locate a point D on AC such that DA = BA. Then draw the segment BD. In the isosceles triangle ∆ABD, 1 ≅ 2. Because mABC = m1+m3, it follows that mABC > m1. Substituting m2 for m1 produces mABC > m2. Because m2 = m3 + mC, m2 > mC. Finally because mABC > m2 ...
