Tommy’s nephew Cody’s birthday
Johnny and Sara were sitting in the school cafeteria when Tommy, a common friend of theirs, joined them. Tommy seemed to be happy and excited. Sara asked Tommy why he was so excited.
Tommy: It’s my nephew Cody’s birthday. The little tyke will be three this Sunday.
Sara: That’s great. You must like him.
Tommy: Yes, I really like the little rascal. Whenever I go to their place, he gives me a big hug and then follows me around until I leave. I guess he not only likes me but also looks up to me. I want to do something special for his birthday.
Johnny: Will you take him out for an ice-cream or something?
Tommy: No. I want to do something bigger and more special than that. Last year I bought him a skate board. Now, he is really good with it. A month ago, we took him to a skate park. He watched boys and girls go onto different types of launch ramps and quarter pikes. He loved it and wanted to try but we didn’t let him because he is too young.
Johnny: So what are you planning?
Tommy wanted to give Cody skate ramps
Tommy: You know Cody’s family lives in the country side and they have a long driveway. I am thinking of buying a couple of small skate ramps for him. I bet he’ll have fun with them but they have to be small enough that he doesn’t hurt himself. He already has a helmet and he will have to wear it too.
Sara: So what’s the problem?
Tommy: If the ramp is very low, it will be no fun for him but if it is too high, he might hurt himself. I saw a ramp with a 10° angle of inclination and the inclined surface on the top was 60 centimeter (cm). That had a height of 10.44 cm. I want another 60 cm ramp with a height of 15 cm. In the beginning, he could use these two ramps separately. When he gets really good, I can stack and glue one ramp on top of the other to make a really high one. I don’t know what the height will be then. Sara, can you help me figure this out?
Sara: Of course, I can help you. First, let us look at the individual ramps.
Ramp A has an incline of 10° and a 60 cm long inclined surface.
Now, sin A = sin 10° = 0.174. That means the height/hypotenuse = 0.174. With a hypotenuse of 60 cm, the height would be 0.174 x 60 cm or 10.44 cm. You can also figure out cos A from this to be √(1 – 0.1742) = 0.985, because sin2 A + cos2 A=1.
Tommy: I already gave you the measurements. Why are you doing this again?
Sara: Just to show you that this relationship works. Now for the Ramp B, the height is 15 cm. That means sin B = 15 cm/60 cm = 0.25 or angle B = 14.48°.
Tommy: I should be able to buy a 15° ramp. That’s good. Okay, sin B = 0.2588 and cos B = 0.966. So this ramp would have a maximum height of 0.2588 x 60 or 15.53 cm.
Stacking skate board ramps to get more height
Sara: If you stack a ramp of incline angle B on of top of a ramp of angle A, the total angle will be A + B. The hypotenuse will remain the same: 60 cm. So we need to find sin (A+B). The simplest way will be to get the value of sine of an angle from a calculator but Johnny here wants to learn some Trig. So I will use one of the identities listed in our Trig book.
Johnny: Thanks Sara. I will go home and study the identities.
Sara: One of the identities is: sin (A + B) = sin A cos B + cos A sin B
For angles A = 10°, B = 15°,
sin A=0.174, cos A= 0.985, sin B=0.2588, cos B = 0.966.
Then sin (A + B) = sin (25°) = 0.174 x 0.966 + 0.985 x 0.2588 = 0.4226.
Tommy: I see the height of the ramp B stacked on top of ramp A will be 60 x 0.4226 or 25.36 cm.
Sara: Is that it then?
Tommy: No. I am worried that 25.36 cm, which is almost ten inches, may be too high for Cody. The guy is not even three feet tall. What height would it be if I just stacked two ramps each of angle A which is 10°.
Sara: You know that sin (A + B) = sin A cos B + cos A sin B. If you were to make the angle A equal to B, it would become: sin (2A) = 2 sin A cos A, Because for angle A which is 10°, sin A = 0.174 and cos A = 0.985, sin (2A) = 0.342 or the height of two 10° ramps which is a 20° incline would be 0.3428 x 60 cm which is approximately 20.6 cm which is a about 8 inches.
Height of a ramp with half the angle
Tommy: Thanks Sara. That sounds perfect but I am going to let Cody’s dad decide. Just tell me one more thing. My brother who is Cody’s dad may decide that this is too risky and I should get the two ramps each with half of the angle 15°. What will be the height of these ramps?
Sara: Here, we can also use the identity: sin (A/2) = ± √(1- cos A)/2.
Because, cos 15° = 0.966, sin A/2 will be 0.13. I did not use the negative height because that would mean a ramp dug to go down into the ground. So the height of each ramp will be 7.8 cm because 60 x 0.13 = 7.8.
Sara: We have to go to a class now. Tommy, tell us what you decide and what your nephew thinks of his birthday present.
Tommy: Thanks Sara. I am so excited.
Johnny and Sara went through their Trig book and found several more identities and tried to prove them.
Challenge: Sara loves to tease her boyfriend Johnny. Today she said to him, “Johnny you think that you are a good bike rider. Let’s say that you are biking on a road which is going down at angle of 15° for 2 km and then it levels off (no incline) for 2 km. After this, it is uphill at 15° for 2 km and then its slope rises by another 7.5° for the last 2 km. What is the total height you would have climbed in the whole bike run?” Base your answer on sin 30° = 0.5.
Solution: There are four parts of the road, each of equal lengths of 2 km. The first has an incline of -15°, the second 0° and the third has an incline of +15°, The increase in heights of the first and the third will cancel out and the second one has no incline. Therefore, the only incline to consider is the fourth one.
The fourth road is an incline of 15° + 7.5° = 22.5°. Given sin 30° = 0.5.
We use the trig identity sin 2 A = 2 sin A cos A as the starting point, to get
cos2 2A = cos2 A – sin2 A, and by substituting B = 2A
cos2 B = cos2 B/2 – sin2 B/2 or sin2 B/2 = (1- cos B)/2 or sin B/2 = ±(√ (1-cos B))/2,
sin 30° = 0.5. Therefore, cos 30° = (√(1 – 0.52))/2 = 0.866.
Therefore, for B= 30°, sin 15° = sin B/2 = ±(√((1- 0.866))/2 = ± 0.2588.
Since 15° is in the first quadrant, sin 15° = + 0.2588.
Now writing C = 2B, we can repeat the above calculations to show that
sin 15°/2 = sin 7.5° = ±(√((1- 0.9659))/2 = ± 0.1305 = + 0.1305 because the angle is in the first quadrant.
The inclination of 22.5° can be calculated as sin of 15° + 7.5° or sin of 30° – 7.5°. Let’s go with the latter. Then sin 22.5° = sin 30° cos 7.5° – sin 7.5° cos 30°.
Because sin 7.5° = 0.1305, cos 7.5° = ±(√((1- 0.1305))/2 = 0.9144.
We already know that sin 30° = 0.5, and cos 30° = 0.866.
Therefore, sin 22.5° = 0.3827.
Therefore, the change in the height due to the fourth part of the road =
2 km x 0.0.3827 = 2000 x 0..3827 meters = 765.4 meters.
Because, the change in the height due to the first three parts is zero, the total change in the height is 765.4 meters.
- Simplify this orally in less than 30 seconds:
sin (170°)/(sin 85° sin 5°)
2 sin 85° cos 85° /sin 85° sin 5° = 2 sin 85° cos 85°/sin 85° cos (90-5°) = 2
- Prove that (1-2cos2x)/(sin x. cos x) = tan x – cot x
We know that sin2 x + cos2 x = 1
Then (1-2cos2x)/(sin x. cos x) = tan x – cot x = (sin2 x + cos2 x -2cos2x)/(sin x. cos x) =
(sin2 x – cos2 x)/(sin x. cos x) = sin2 x/ (sin x .cos x) – cos2 x / sin x.cos x = sin x/cos x – cos x /sin x = tan x – cot x