Finding a point on a line (from drawing of house on hill)

[maxthebat12]

Hi,

I am struggling on how to find point A on this diagram given to me from my high school teacher.

Could I please get some assistance.

 

[mmm4444bot]

I am struggling on how to find point A on this diagram given to me from my high school teacher.
View attachment 8624

Are you supposed to report A's location using xy-coordinates? If so, are we supposed to assume that the horizontal line segment at the bottom lies on the x-axis and that the Origin is located at its left end?

 

[maxthebat12]

Are you supposed to report A's location using xy-coordinates? If so, are we supposed to assume that the horizontal line segment at the bottom lies on the x-axis and that the Origin is located at its left end?

Hi thanks for reviewing my question!

I was just given just these information. But the distance from 27 to 25 is 100 horizontally and from 27 to the first wall of the house is 20.

 

[mmm4444bot]

Well, if your teacher handed you the diagram (with nothing else) and simply said, "Find point A", before walking away, then I would guess that you're free to answer the question in any form that makes sense. Do you agree?

What has your class been studying, lately?

Have you learned about any of the following topics, yet?

Alternate Interior Angles

Similar right triangles

Trigonometry

Slope Formula

Slope-Intercept Form

Point-Slope Form

 

[maxthebat12]

Well, if your teacher handed you the diagram (with nothing else) and simply said, "Find point A", before walking away, then I would guess that you're free to answer the question in any form that makes sense. Do you agree?

What has your class been studying, lately?

Have you learned about any of the following topics, yet?

Alternate Interior Angles

Similar right triangles

Trigonometry

Slope Formula

Slope-Intercept Form

Point-Slope Form

Is there a way to solve this through trigonometry?

 

[mmm4444bot]

Is there a way to solve this through trigonometry?

Yup, but that approach would be my last choice because of extra steps.

A more direct approach would be to set up and solve a proportion, based on similar right triangles.

In the diagram below, the green lines are parallel. What can you say about the angles marked in red?

 

[maxthebat12]

Yup, but that approach would be my last choice because of extra steps.

A more direct approach would be to set up and solve a proportion, based on similar right triangles.

In the diagram below, the green lines are parallel. What can you say about the angles marked in red?

View attachment 8626

Those angles will always be the same since the lines are parallel?

 

[maxthebat12]

Hmmmm....what do "25" and "27" represent?

Teacher wants us to use two points to find a point along that line (say 27 is the top of a hill and 25 is the bottom)

I'd assume 27m and 25m with a distance of 100m from the two points and 20m from the 27 point to the first wall of a house.

 

[mmm4444bot]

Those angles will always be the same since the lines are parallel?

Yes -- alternate interior angles are always equal, when a transversal crosses a pair of parallel lines.

Form two right triangles. One has base 100, and the other has base 20. As each of these right triangles contains the same acute angle, they must be similar right triangles.

You have enough information to calculate the hypotenuse of the larger triangle. The hypotenuse of the smaller triangle is what you need to find. (This allows you to answer the question by stating the distance that point A is located from the top of the hill, as measured down the incline.)

Because the two right triangles are similar, their sides are proportional. Set up a proportion.

 

[mmm4444bot]

Can you post EXACTLY your teacher's original problem?

maxthebat12 has already stated that the uploaded diagram is what they received and that nothing else was provided.

I'm thinking that the given numbers represent linear measurements, as no angle units are shown.

 

[mmm4444bot]

[Trigonometry] would be my last choice …

On second thought, the trigonometric approach is not a lot of extra work; I had failed to consider using the arctangent function (to get the measure of the angle marked in red), when I posted last night.