Right angle trigonometry homework question

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SUMMARY

The discussion focuses on solving a right angle trigonometry problem involving a steel plate shaped as one-fourth of a circle with a radius of 60 centimeters. The key challenge is determining the coordinates of two holes drilled at a radial distance of 56 centimeters from the center, positioned at angles of 30° and 60° from the horizontal. The hypotenuse of 56 centimeters is derived from the radius of the circle, and the solution involves converting polar coordinates to rectangular coordinates using trigonometric functions.

PREREQUISITES
  • Understanding of right angle trigonometry, specifically 30-60-90 triangles.
  • Knowledge of polar and rectangular coordinate systems.
  • Familiarity with sine and cosine functions for calculating coordinates.
  • Basic geometry concepts related to circles and radii.
NEXT STEPS
  • Study the properties of 30-60-90 triangles and their side ratios.
  • Learn how to convert polar coordinates to rectangular coordinates.
  • Practice problems involving trigonometric functions in real-world applications.
  • Explore the use of trigonometric identities in solving geometric problems.
USEFUL FOR

Students studying trigonometry, educators teaching geometry, and anyone involved in practical applications of mathematics in engineering or design.

xxwinexx
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Homework Statement


A steel plate has the form of one-fourth of a circle with a radius of 60 centimeters. Two two-centimeter holes are to be drilled in the plate positioned as shown in the figure. Find the coordinates of the center of each hole.

Homework Equations


I know it's got to be a simple sin/cos right angle equation, but I've no clue how they came about to getting the 56 centimeter measurement to be the hypotenuse on the solution that I have in my manual.

The Attempt at a Solution



I attempted drawing a few right angle triangles that I thought would work, but nothing came from it. If someone could explain how the solution manual was able to get a triangle with 56 as the hypotenuse, I would be able to move forward from there.

I've attached both the image of the problem, and the image of the first part of the solution worked out.
 

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for the first hole you have a 30-60-90 rt triangle with hyp 56 so you should be able to compute the x1 and y1

similarly for the x2 y2
 
It's because it's the radius of the circle. Every hypotenuse along the circumference of a circle that is measured from the centre is going to be the radius of the circle.
 
xxwinexx said:

Homework Statement


A steel plate has the form of one-fourth of a circle with a radius of 60 centimeters. Two two-centimeter holes are to be drilled in the plate positioned as shown in the figure. Find the coordinates of the center of each hole.


Homework Equations


I know it's got to be a simple sin/cos right angle equation, but I've no clue how they came about to getting the 56 centimeter measurement to be the hypotenuse on the solution that I have in my manual.
It looks to me like the two drilled holes are a radial distance of 56 cm from the center, and at angles of 30° and 60° from the horizontal edge.

These values aren't calculated - they're part of the given information in the problem. The positions of the two holes are essentially in polar coordinates, and your job is to find the rectangular coordinates of the holes.
xxwinexx said:

The Attempt at a Solution



I attempted drawing a few right angle triangles that I thought would work, but nothing came from it. If someone could explain how the solution manual was able to get a triangle with 56 as the hypotenuse, I would be able to move forward from there.

I've attached both the image of the problem, and the image of the first part of the solution worked out.
 
jedishrfu said:
for the first hole you have a 30-60-90 rt triangle with hyp 56 so you should be able to compute the x1 and y1

similarly for the x2 y2

Right, that's basically what the solution is saying, I guess I just can't see how they figured that right triangle/hypotenuse out..

Edit: Ohhhh...I wasn't thinking of the 56 as a radial measurement. I feel really dumb now. Thanks guys!
 

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