Chapter 7: right triangles and trigonometry Rachel Jeong Period 5 Laptop Honors Geometry

7-5: angles of elevation and depression An angle of elevation is the angle between the line of sight and the horizontal when an observer looks upward An angle of depression is the angle between the line of sight when on observer looks downward and the horizontally. Find the angle of elevation of the sun when a 7.6-meter flagpole casts a meter shadow. Round to the nearest tenth of a degree.

Example 2 for Angle of elevation and depression Example2 AD||CE, so ECD is congruent to ADC and ECB is congruent to ABC as they are the alternate interior angles. So, mADC = mECD = 8 and mABC= mECB = 11. In the right triangle ABC with right angle at A, Cross multiply: AB Tan11˚=60. Divide both sides by tan 11˚. AB=60/tan11˚ AB In the right triangle ADC with right angle at A Cross multiply: AD tan8˚=60 Divide both sides by tan 8. AD=60/tan8˚ AD Three point A, B and C are collinear and B is in between the points A and D. So, BD = AD – AB. BD≈ – BD≈ yards So, the distance between the merry–go–around and the Ferris wheel is about yards.

Example 3 for Angle of elevation and depression Example 3 The situation is sketched as shown in the figure. Kirk's eyes are at point B and the geyser reaches the point C. We have to find the angle of elevation to the top of spray. The angle of elevation is the angle between the line of sight and the horizontal when a person looks upward. So, here the angle of elevation is mCBD In the right triangle BCD with right angle at D, TanCBD=CD/BD We know CE and AE. We have to find CD and BD. Because AE and BD are horizontal lines, AE||BD. Also the lines AB, DE are perpendicular to the horizontal lines. So, the quadrilateral formed is a rectangle. We have DE = AB = 6 and BD = AE = 200. Three points C,D and E are on a straight line and so we have CD = CE – DE. CD = 175 – 6 =169

7-3 special right triangles In triangle, the length of the hypotenuse is √2 times the length of a leg In a triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is √3 times the length of the shorter leg. Example1 The perimeter of an equilateral triangle is three times the measure of a side. Perimeter = 3x, if the measure of the side is x. So, 45 = 3x. Divide by 3 on both the sides. x = 45/3 = 15 cm So, the measure of the side is 15 cm The altitude of an equilateral triangle divides the triangle into two congruent 30–60–90 triangles. The length of the altitude is the length of the longest leg in the 30–60–90 triangle.

Example2 for special right triangle Example 2 Given that the mDHB = 60. Also the mBDH = 90. So, the triangle BDH is a 30–60–90 triangle. The side opposite to larger angle is longer leg. So, BDis the longer leg and DHs the shorter leg. From Theorem 7.7, in a 30–60–90 triangle, the length of the longer leg is √ 3 times the length of the shorter leg. Substitute BD: From Theorem 7.7, in a 30–60–90 triangle, the length of the hypotenuse is twice the length of the shorter leg. So, BH=2(DH) Substitute DH BH=2(8) BC=16

Example3 for special right triangle Example 3 From the Theorem 7.6, the length of the hypotenuse is √ 2 times the length of a leg in 45– 45–90 triangle.

7-1 Geometric means Geometric mean: for two positive numbers a and b, the geometric mean is the positive number x where the proportion a:x=x:b is true. This proportion can be written using fractions as a/x=x/b or with cross products as x²=ab or x=√ab Theorem 7.1: if the altitude is drawn from the vertex of the right angle of a right triangle to its hypotenuse, then the two triangles formed are similar to the given triangle and to each other Theorem 7.2: the measure of an altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse Theorem 7.3: if the altitude is drawn from the vertex of the right angle of a right triangle to its hypotenuse, then the measure of a leg of the triangle is the geometric mean between the measures of the hypotenuse and the segment of the of the hypotenuse adjacent to that leg Example 1: Applying Theorem 7.3 for the leg BD Since AC = 10, BC = 6 and CD = x, the above fraction can be written as: Find the cross products. 10x = 6 · 6 10x = 36 Dividing by 10 on both sides: x = 36/10 = 3.6 So, CD = x = 3.6. From the figure, x + y = y = 10 y = 10 – 3.6 = 6.4 We get the values as: x = 3.6 and y = 6.4

Example 2 for Geometric mean Example 2 Given: BD = 5 and CD = 9 Let AD be x. Find the cross products. x ·x = 5 · 9 x 2 = 45 Take the positive square root of each side. Simplifying, we get: x ≈6.7 So, AD is about 6.7.

Example 3 for Geometric mean Example 3 Applying Theorem 7.3 for the leg BC Since AC=15, BC=5, and CD=y, the above fraction can be written as Find the cross products. 15 ·y = 5 · 5 15y = 25 y = 25/15 y = 5/3 Given: x + y = 15 Substitute the value of y as 5/3. x + 5/3 = 15 x = 15 – 5/3 x = 40/3 Find the cross products. z z = 15 (40/3) z 2 = 5 · 40 z 2 = 200