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7.5 & 7.6– Apply the Sin-Cos-Tan Ratios. Hypotenuse: Opposite side: Adjacent side: Side opposite the reference angle Side opposite the right angle Side.

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Presentation on theme: "7.5 & 7.6– Apply the Sin-Cos-Tan Ratios. Hypotenuse: Opposite side: Adjacent side: Side opposite the reference angle Side opposite the right angle Side."— Presentation transcript:

1 7.5 & 7.6– Apply the Sin-Cos-Tan Ratios

2 Hypotenuse: Opposite side: Adjacent side: Side opposite the reference angle Side opposite the right angle Side touching the reference angle, but not the hypotenuse A hypotenuse opposite adjacent

3 a b c A B C sin cos tan sin A° = opp hyp cos A° = adj hyp tan A° = opp adj sin A° = acac cos A° = bcbc tan A° = abab SOHCAHTOA Note: You cannot use trig ratios on the right angle

4 1. Find the sin R. Write your answer as a fraction. sin R° = 15 17 SOH – CAH – TOA A H O

5 2. Find the cos A. Write your answer as a fraction. cos A° = 16 20 SOH – CAH – TOA A H O

6 3. Find the tan B. Write your answer as a fraction. tan B° = 24. 10 SOH – CAH – TOA A H O

7 4. Write the trigonometric ratio as a decimal to two decimal places. a. sin 62° b. cos 25°c. tan 38° 0.880.91 0.78

8 5. Solve for x. Round to two decimal places. 1 x = 5  sin 24° x =x = 2.03 a.

9 5. Solve for x. Round to two decimal places. 1 x  tan35° = 4 tan35° x = 5.71 b.

10 Steps to solving with trig: 1. 2. 3. 4. Label the sides with the reference angle Choose a side to solve for What side do you know and what side are you solving for? Identify which trig ratio to use

11 cos 50° = a. 10 10  cos 50° = a 6.43 = a A H 6. Approximate the missing side length to two decimal places. SOH – CAH – TOA 1 O

12 sin 42° = c. 7 7  sin 42° = c 4.68 = c SOH – CAH – TOA 1 A H O 6. Approximate the missing side length to two decimal places.

13 mn tan 39° = 25 m m  tan 39° = 25 m = 30.87 SOH – CAH – TOA 1 sin 39° = 25 n n  sin 39° = 25 n = 39.73 1 tan 39° sin 39° A H O 7. Approximate the missing side lengths to two decimal places.

14 tu sin 51° = t. 32 32  sin 51° = t 24.87 = t SOH – CAH – TOA 1 cos 51° = u. 32 32  cos 51° = u 20.14 = u 1 A H O 7. Approximate the missing side lengths to two decimal places.

15 ab tan 57° = 14 a a  tan 57° = 14 SOH – CAH – TOA 1 tan 57° a = 9.09 sin 57° = 14 b b  sin 57° = 14 1 sin 57° b = 16.69 A H O 7. Approximate the missing side lengths to two decimal places.

16 mn cos 31° = 79 m m  cos 31° = 79 SOH – CAH – TOA 1 cos 31° m = 92.16 tan 31° = n. 79 79  tan 31° = n 1 n = 47.47 A H O 7. Approximate the missing side lengths to two decimal places.


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