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**ANGLE ANGULAR MEASUREMENT**

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**An angle has “a positive” sign if its rotation is anticlockwise **

Terminal side Initial side Angle is defined a rotation result from initial side to the terminal side An angle has “a positive” sign if its rotation is anticlockwise An angle has “a negative” sign if its rotation is clockwise We only talk about the magnitude of angle~ not observe its signs Terminal side

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**What is a radian? radius arc=radius o Angle = 1 rad radius**

Why do mathematicians use radians instead of degrees?

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**How many times does the radius divide into the circumference?**

There are 2 radians in a circle. 1 radian = = 57.3o

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**THE TRIGONOMETRIC RATIOS**

BET1542 ENG. MATHEMATICS II THE TRIGONOMETRIC RATIOS 4 SEMESTER I SESSION 2008/09

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**Convert each angle in radians to degrees.**

Convert each angle in degrees to radians. 1. 2c 2. 5c c c c 1. 65o o o o o 1.13c 114.6o 3.49c 286.5o 540o 90o 240o

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**Two important formula using radians**

Length of an arc using radians Area of a sector using radians

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Trigonometric Ratios

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**Objectives/Assignment**

Find the since, the cosine, and the tangent of an acute triangle. Use trionometric ratios to solve real-life problems

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Finding Trig Ratios A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. The word trigonometry is derived from the ancient Greek language and means measurement of triangles. The three basic trigonometric ratios are sin, cos, and tan

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Trigonometric Ratios Let ∆ABC be a right triangle. The since, the cosine, and the tangent of the acute angle A are defined as follows. Side adjacent to A b cos A = = hypotenuse c Side opposite A a sin A = = hypotenuse c Side opposite A a tan A = = Side adjacent to A b

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Exercise: In the PQR triangle is right angled at R happens cos P = 8/ 17…find the value of tg P and tg Q ?

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Exercises 1. PQR triangle is right angled at R & sin P cosQ = ⅗. Find out the value c ABC triangle is right angle at A. BC=p, AD is perpendicular at BC, DE perpendicular at AC, angle B = Q, prove that : DE=psin2 Qcos Q 2 D E ABC triangle is rightangle at A. BC=p, AD is perpendicular at BC, DE perpendicular at AC, angle B = Q, prove that : DE=psin2 Qcos Q A B

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**Evaluating Trigonometric Functions**

Acute angle A is drawn in standard position as shown. Right-Triangle-Based Definitions of Trigonometric Functions For any acute angle A in standard position,

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**Finding Trigonometric Function Values**

Finding Trigonometric Function Values of an Acute Angle in a Right Triangle Example Find the values of sin A, cos A, and tan A in the right triangle. Solution length of side opposite angle A is 7 length of side adjacent angle A is 24 length of hypotenuse is 25

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**Trigonometric Function Values of Special Angles**

Angles that deserve special study are 30º, 45º, and 60º. Using the figures above, we have the exact values of the special angles summarized in the table on the right.

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**Let’s try to prove it ! OA=OB OA2 + OB2 = OC2 OA2 + OA2 = r2 2OA2 = 1**

X 45O OA=OB OA2 + OB2 = OC2 OA2 + OA2 = r2 2OA2 = 1 OA2 = ½ = OB How about 300 , 600 and 900 OA = How about 300 , 600 and 900

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**Cofunction Identities**

In a right triangle ABC, with right angle C, the acute angles A and B are complementary. Since angles A and B are complementary, and sin A = cos B, the functions sine and cosine are called cofunctions. Similarly for secant and cosecant, and tangent and cotangent.

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**Cofunction Identities**

If A is an acute angle measured in degrees, then If A is an acute angle measured in radians, then Note These identities actually apply to all angles (not just acute angles).

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Reference Angles A reference angle for an angle , written , is the positive acute angle made by the terminal side of angle and the x-axis. Example Find the reference angle for each angle. 218º (b) Solution (a) = 218º – 180º = 38º (b)

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**Special Angles as Reference Angles**

Example Find the values of the trigonometric functions for 210º. Solution The reference angle for 210º is 210º – 180º = 30º. Choose point P on the terminal side so that the distance from the origin to P is 2. A 30º - 60º right triangle is formed.

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**Finding Trigonometric Function Values Using Reference Angles**

Example Find the exact value of each expression. cos(–240º) (b) tan 675º Solution –240º is coterminal with 120º. The reference angle is 180º – 120º = 60º. Since –240º lies in quadrant II, the cos(–240º) is negative. Similarly, tan 675º = tan 315º = –tan 45º = –1.

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Finding Angle Measure Example Find all values of , if is in the interval [0º, 360º) and Solution Since cosine is negative, must lie in either quadrant II or III. Since So the reference angle = 45º. The quadrant II angle = 180º – 45º = 135º, and the quadrant III angle = 180º + 45º = 225º.

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**Ex. 1: Finding Trig Ratios**

Compare the sine, the cosine, and the tangent ratios for A in each triangle beside. By the Similarity Theorem, the triangles are similar. Their corresponding sides are in proportion which implies that the trigonometric ratios for A in each triangle are the same.

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**Ex. 1: Finding Trig Ratios**

Large Small opposite 8 sin A = ≈ 0.4706 4 ≈ 0.4706 hypotenuse 17 8.5 adjacent 7.5 cosA = 15 ≈ 0.8824 ≈ 0.8824 hypotenuse 8.5 17 opposite tanA = 8 4 ≈ 0.5333 ≈ 0.5333 adjacent 15 7.5 Trig ratios are often expressed as decimal approximations.

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**Ex. 2: Finding Trig Ratios**

opposite 5 sin S = ≈ 0.3846 hypotenuse 13 adjacent cosS = 12 ≈ 0.9231 hypotenuse 13 opposite tanS = 5 ≈ 0.4167 adjacent 12

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**Ex. 2: Finding Trig Ratios—Find the sine, the cosine, and the tangent of the indicated angle.**

opposite 12 sin S = ≈ 0.9231 hypotenuse 13 adjacent cosS = 5 ≈ 0.3846 hypotenuse 13 opposite tanS = 12 ≈ 2.4 adjacent 5

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Notes: If you look back, you will notice that the sine or the cosine of an acute triangles is always less than 1. The reason is that these trigonometric ratios involve the ratio of a leg of a right triangle to the hypotenuse. The length of a leg or a right triangle is always less than the length of its hypotenuse, so the ratio of these lengths is always less than one.

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**Using Trigonometric Ratios in Real-life**

Suppose you stand and look up at a point in the distance. Maybe you are looking up at the top of a tree as in Example 6. The angle that your line of sight makes with a line drawn horizontally is called angle of elevation.

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**Ex. 6: Indirect Measurement**

You are measuring the height of a Sitka spruce tree in Alaska. You stand 45 feet from the base of the tree. You measure the angle of elevation from a point on the ground to the top of the top of the tree to be 59°. To estimate the height of the tree, you can write a trigonometric ratio that involves the height h and the known length of 45 feet.

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**The math The tree is about 76 feet tall. Write the ratio**

tan 59° = opposite adjacent Write the ratio tan 59° = h 45 Substitute values Multiply each side by 45 45 tan 59° = h Use a calculator or table to find tan 59° 45 (1.6643) ≈ h Simplify 75.9 ≈ h The tree is about 76 feet tall.

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**Ex. 7: Estimating Distance**

Escalators. The escalator at the Wilshire/Vermont Metro Rail Station in Los Angeles rises 76 feet at a 30° angle. To find the distance d a person travels on the escalator stairs, you can write a trigonometric ratio that involves the hypotenuse and the known leg of 76 feet. 30°

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Now the math sin 30° = opposite hypotenuse 30° Write the ratio for sine of 30° sin 30° = 76 d Substitute values. d sin 30° = 76 Multiply each side by d. sin 30° 76 d = Divide each side by sin 30° 0.5 76 d = Substitute 0.5 for sin 30° d = 152 Simplify A person travels 152 feet on the escalator stairs.

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Lesson 9.5 Trigonometric Ratio

Lesson 9.5 Trigonometric Ratio

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