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# • A six-meter-long ladder leans against a building. If the l

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## Janus Laxa

on 13 March 2014

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#### Transcript of • A six-meter-long ladder leans against a building. If the l

• A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the ground, how far up the wall does the ladder reach? How far from the wall is the base of the ladder? Round your answers to two decimal places, as needed.
First, I'll draw a picture. It doesn't have to be "good"; it just needs to be clear enough that I can keep track of what I'm doing. My picture is at right:
I need to find the height h. Since they've given me an angle measure and "opposite" and the hypotenuse for this angle, I'll use the sine ratio for finding the height:
sin(60°) = h/6
6×sin(60°) = h = 3sqrt[3]
Plugging this into my calculator, I get an approximate value of 5.196152423.

For the base, I'll use the cosine ratio:
cos(60°) = b/6
6×cos(60°) = b = 3
and its base is 3 meters from the wall.
Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting. For instance, if they hadn't told me to round in the exercise above, my value for the height should have been the value with the radical.
A five-meter-long ladder leans against a wall, with the top of the ladder being four meters above the ground. What is the approximate angle that the ladder makes with the ground?
They've given me the "opposite" and the hypotenuse, and asked me for the angle value. For this, I'll need to use inverse trig ratios.
sin(a) = 4/5
a = sin–1(4/5) = 53.13010235...

You use a transit to measure the angle of the sun in the sky; the sun fills 34' of arc. Assuming the sun is 92,919,800 miles away, find the diameter of the sun. Round your answer to the nearest mile.
The two lines along the side of my triangle measure the lines of sight from Earth to the sides of the Sun. What if I add another line, being the direct line from Earth to the center of the Sun?

Now that I've got this added line, I have a right triangle; two right triangles, actually, but I only need one.
Because the Sun is so far away, obviously this picture is not "to scale", and I'll have to ignore the fact that the distance measured doesn't quite match the picture. But this should be good enough.

(The angle measure, "thirty-four arc minutes", is equal to 34/60 degrees. Dividing this in half is how I got 17/60 of a degree for the smaller angle.)
I need to find the width of the Sun. That width will be twice the base of one of the right triangles. With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio:
tan(17/60°) = b/92919800
92919800×tan(17/60°) = b = 459501.4065...
This is just half the width; carrying the calculations in my calculator (to minimize round-off error), I get a value of 919002.8129. This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise.

The diameter is about 919,003 miles.
A private plane flies 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point? What is its bearing from that starting point?
bearings tell me the angles from "due north", in a clockwise direction. Since 130 – 40 = 90, these two bearings will give me a right triangle. From the times and rates, I can find the distances:
1.3 × 110 = 143
1.5 × 110 = 165
Now that I have the lengths of the two legs, I can set up a triangle:

I can find the distance by using the Pythagorean Theorem:
1432 + 1652 = c2
20449 + 27225 = c2
47674 = c2
c = 218.3437657...
The 165 is opposite the unknown angle, and the 143 is adjacent, so I'll use the inverse of the tangent ratio to find the angle's measure:
165/143 = tan(θ)
tan–1(165/143) = θ = 49.08561678...
But this is not the "bearing", since the bearing is the angle with respect to "due north". I need to add in the original forty-degree angle to get my answer:

The plane is about 218 miles away, at a bearing of about 89°.
A ramp so that people in wheelchairs can access a building. If the ramp must have a height of 8 feet, and the angle of the ramp must be about 5°, how long must the ramp be?
Given that we know the angle of the ramp and the length of the side opposite the angle, we can use the sine ratio to find the length of the ramp, which is the hypotenuse of the triangle:

sin(5º) = 8/L
Lsin(5º) = 8
L = 8/ sin(5º)
L = 91.8 ft

This may seem like a long ramp, but in fact a 5° angle ramp. This explains why many ramps are comprised of several sections, or have turns. The additional distance is needed to make up for the small slope.

Three towns, Airdrie, Okotoks and Cochrane are located so that Airidrie is 63 km rom Okotoks and 34 km from Cochrane. If the angle at Airdrie is 73º, then how far is Okotoks from Cochrane?
a = b² c² - 3bc cosA
= (34km)² + (63km)² - 2(34km)(63km) cos 73º
= 5125km - 4284km cos73º
= 841km cos73º
a = 245.88km

A golfer hots her drive 200m at an angle of 8º to the left of the straight line path to the flag. Her second shot travels 45m straight at the flag and rolls in! What was the total length of the hole from the tee box to the flag?
sin C/ c = sin A/ a
sin C = sin 8º/45m (200m)
C = 38º12'36.41"

B = 180 - (A + C)
= 180º - (8º + 38º12'36.41")
B = 133º47'23.59"

b/ sin B = a/ sin A
b = 45m/ sin 8º (sin 133º47'23.59")
b = 233.41m
The sides of a triangular grass lot measures 72m, 60, and 65m respectively. What are the angles of the grass field?
cos A = b² + c² - a²/ 2bc
cos A =5809/9360
A = 51º 39'15.16"

sin B/ b = sin A/ a
sin B = sin 51º 39'15.16"/ 60 (65)
B = 58º10'20.17"

C = 180º - (A + B)
C = 70º10'24.67"
Side AB measures 38ft, the angle opposite to the side is 20º and A is 80º., Solve for the triangle.
a/ sin A = b/ sin B
a = 38º/ sin 20º (sin 80º)
a = 109.42 ft

C = 180º - (80º + 20º)
C = 80º

c/ sin C = b/ sin B
c = 38ft/ sin 20 (sin 80º)
c = 109.42 ft
Anna's distance from Mike is 25ft. Mike is 3ft away from Jack. The angle where Anna is the vertex is 20º. While the angle where Mike is the vertex is 30º. What is the measure of the angle where Jack is the vertex and how far is Anna from Jack.
A = 180º - (B + C)
A = 180º - (20º + 30º)
A = 120º

c/sin C = a/ sin A
c = 25 ft/ sin 120º (sin 30º)
c = 14.43 ft
Right Triangles
Oblique Triangles
Group Members:
* Rafael Caingcoy
* Francis Del Mundo
* Janus Mikhael Laxa
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