Math 1330 Section 6.2 1 Section 7.1: Right-Triangle Applications in This Section, We'll Solve Right Triangles. in Some Proble

Math 1330 Section 6.2 1 Section 7.1: Right-Triangle Applications in This Section, We'll Solve Right Triangles. in Some Proble

Math 1330 Section 6.2 Section 7.1: Right-Triangle Applications In this section, we’ll solve right triangles. In some problems you will be asked to find one or two specific pieces of information, but often you’ll be asked to “solve the triangle,” which means that you will find all lengths and measures that were not given. You’ll use the six trigonometric functions of an angle to do this. In some cases, you will be able to use properties of the 30 ° − 60 ° − 90 °triangles or of 45 ° − 45 ° − 90 ° . Example 1: Calculate x and y from the figure below. y x 100cm 30 o Example 2: In ∆ABC with right angle C, ∠A = 46 ° and AC = 12. Find BC. Round to the answer to the nearest hundredth. Example 3: Find x and y in the triangle below. y 10 ft. 40 ° x 1 Math 1330 Section 6.2 Example 4: Draw a diagram to represent the given situation. Then find the indicated measures to the nearest degree. An isosceles triangle has sides measuring 20 inches, 54 inches and 54 inches. What are the measures of the angle? Angle of Elevation The angle of elevation is an angle that s formed by the horizontal ray and another ray above the horizontal. So when viewing an object at a point above the horizontal, the angle between the line of sight and horizontal is angle of elevation in the figure below. Example 5: Draw a diagram to represent the given situation. The find the indicate measure to the nearest tenth. The angle elevation to the top of a building from a point on the ground 125 feet away from the building is 8° . How tall is the building? 2 Math 1330 Section 6.2 Example 6: Draw a diagram to represent the given situation. The find the indicate measure to the nearest tenth. A 16-foot ladder leans against a building. The ladder forms an angle of 70 °with the ground. a) How high up the building does the ladder reach? b) What is the horizontal distance from the foot of the ladder to the base of the building? Angle of Depression The angle of depression is an angle that s formed by the horizontal ray and another ray below the horizontal. So when viewing an object at a point above the horizontal, the angle between the line of sight and horizontal is angle of elevation in the figure below. Example 7: Draw a diagram to represent the given situation. The find the indicate measure to the nearest tenth. Dave is at the top of a hill. He looks down and spots his car at a 61 °angle of depression. If the hill is 59 meters high, how far is his car from the base of the hill? 3 Math 1330 Section 6.2 Example 8: Mike stands 450 feet from the base of the Empire State Building and sights the top of the building. If the Empire State Building is 1,453 feet tall, approximate the angle of elevation from Mike’s perspective as he sights the top of the building. (Disregard Mike’s height in your calculations.) 4 Math 1330 Section 7.2 7.2 Area of a Triangle In this section, we’ll use a familiar formula and a new formula to find the area of a triangle. Given a b 1 Recall the area of a triangle is given by A = bh . We normally used this formula when we knew 2 the height of the triangle. However if the height is not given we must solve for it some way. So we solve for the height of the triangle by breaking the given triangle down into right triangles and use our trigonometric function (sine) to solve for the height. Area of a Trangle 1 A = ab sin θ 2 a, b are the lengths of two sides of a triangle θ is the angle between them. Example 1: Find the exact area of the triangle. 30 ° 11 Math 1330 Section 7.2 Example 2: Find the exact area of the triangle. 29 120 ° 17 Example 3: Find the area of an isosceles triangle with legs measuring 12 inches and the base angle is 30 degrees each. Example 4: If the area of ∆EFG is 30 in 2, e = 8 in. and g = 15 in, find all possible measures of angle F. Math 1330 Section 7.2 Example 5: A regular hexagon is inscribed in a circle of radius 6m. Find the area of the hexagon. Area of a Segment of a Circle You can also find the area of a segment of a circle. The shaded area of the picture is an example of a segment of a circle. B O A To find the area of a segment, find the area of the sector with central angle θ and radius OA. Then find the area of ∆OAB . Then subtract the area of the triangle from the area of the sector. 1 Area of a sector of a circle = r 2θ 2 − Area of the segment = Area sec tor Area triangle Example 6: Find the area of the segment of the circle with radius 6 meters and central angle 3π measuring . 4 Math 1330 Section 7.3 Section 7.3: Law of Sines and Law of Cosines Sometimes you will need to solve a triangle that is not a right triangle. This type of triangle is called an oblique triangle. To solve an oblique triangle you will not be able to use right triangle trigonometry. Instead, you will use the Law of Sines and/or the Law of Cosines. You will typically be given three parts of the triangle and you will be asked to find the other three. The approach you will take to the problem will depend on the information that is given. If you are given SSS (the lengths of all three sides) or SAS (the lengths of two sides and the measure of the included angle), you will use the Law of Cosines to solve the triangle. If you are given SAA (the measures of two angles and one side) or SSA (the measures of two sides and the measure of an angle that is not the included angle), you will use the Law of Sines to solve the triangle. Recall from your geometry course that SSA does not necessarily determine a triangle. We will need to take special care when this is the given information. Here’s the Law of Cosines. In any triangle ABC, = + − 2 cos = + − 2 cos = + − 2 cos The development of one case of this formula is given in detail in the online text. Here’s the Law of Sines. In any triangle ABC, sin sin sin = = The development of this formula is given in detail in the online text. Here are some facts about solving triangles that may be helpful in this section. If you are given SSS, SAS or SAA, the information determines a unique triangle. If you are given SSA, the information given may determine 0, 1 or 2 triangles. If this is the information you are given, you will have some additional work to do. Since you will have three pieces of information to find when solving a triangle, it is possible for you to use both the Law of Sines and the Law of Cosines in the same problem. When drawing a triangle, the measure of the largest angle is opposite the longest side; the measure of the middle-sized angle is opposite the middle-sized side; and the measure of the smallest angle is opposite the shortest side. Suppose a, b and c are suggested to be the lengths of the three sides of a triangle. Suppose that c is the biggest of the three measures. In order for a, b and c to form a triangle, this inequality must be true: a + b > c . So, the sum of the two smaller sides must be greater than the third side. 1 .

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