Tricky triangles word problems, can't seem to get the right answer.

J

James Ibarra

New member
Joined
Apr 22, 2020
Messages
26
word problems:

1. A man standing on a 48.5 meter building high, has an eyesight height of 1.5 meters from the top of the building, took a depression reading of the top of another nearby building and nearest wall, which are 50 degrees and 80 degrees respectively. Find the height of the nearby building in meters. The man is standing at the edge of the building and both buildings lie on the same horizontal plane. ANSWER: 39.49 meters


2.If an equilateral triangle is circumscribed about a circle of radius 10 cm, determine the measurement of the side of the triangle. ANSWER: 34.64 centimeters


3.A and B are summits of two mountains that rise from a horizontal plain. B being 1200 meters above the plain. Find the height of A, it being given that its angle of elevation as seen from a point C in the plain (in the same vertical plane with A and B) is 50 degrees, while the angle of depression of C viewed from B is 28 degrees 58 minutes and the angle subtended at B by AC is 50 degrees.ANSWER 1537.493 meters


4.The sides of a triangular lot are 130 meters, 180 meters and 190 meters. The lot is to be divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line. ANSWER: 125 meters


5.Points A and B 1000 m apart are plotted on a straight highway running East and West. From A, the bearing of a tower C is 32 degrees W of N and from B the bearing of C is 26 degrees N of E. Approximate the shortest distance of tower C to the highway. ANSWER: 374 meters





These question were given to us as brain teaser not as homework to pass during the pandermic. The questions where surely teasing my tiny brain as I cannot get the right answer. I thought it would be easy because the answer was already given and we only need to provide the solution. I didn't even know triangles could get this tricker and confusing. BTW THANKS FOR EVERY PERSON THAT NOTICES THIS. and shout out to mark of this forum for helping me understand my trigo identities homeowrk. thanks!
 
Last edited by a moderator:
James Ibarra said:
word problems:

1. A man standing on a 48.5 meter building high, has an eyesight height of 1.5 meters from the top of the building, took a depression reading of the top of another nearby building and nearest wall, which are 50 degrees and 80 degrees respectively. Find the height of the nearby building in meters. The man is standing at the edge of the building and both buildings lie on the same horizontal plane. ANSWER: 39.49 meters


2.If an equilateral triangle is circumscribed about a circle of radius 10 cm, determine the measurement of the side of the triangle. ANSWER: 34.64 centimeters


3.A and B are summits of two mountains that rise from a horizontal plain. B being 1200 meters above the plain. Find the height of A, it being given that its angle of elevation as seen from a point C in the plain (in the same vertical plane with A and B) is 50 degrees, while the angle of depression of C viewed from B is 28 degrees 58 minutes and the angle subtended at B by AC is 50 degrees.ANSWER 1537.493 meters


4.The sides of a triangular lot are 130 meters, 180 meters and 190 meters. The lot is to be divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line. ANSWER: 125 meters


5.Points A and B 1000 m apart are plotted on a straight highway running East and West. From A, the bearing of a tower C is 32 degrees W of N and from B the bearing of C is 26 degrees N of E. Approximate the shortest distance of tower C to the highway. ANSWER: 374 meters





These question were given to us as brain teaser not as homework to pass during the pandermic. The questions where surely teasing my tiny brain as I cannot get the right answer. I thought it would be easy because the answer was already given and we only need to provide the solution. I didn't even know triangles could get this tricker and confusing. BTW THANKS FOR EVERY PERSON THAT NOTICES THIS. and shout out to mark of this forum for helping me understand my trigo identities homeowrk. thanks!
Click to expand...
Mr. Ibarra,

You have been told in your last post - one problem/thread.
 
James Ibarra said:
word problems:

1. A man standing on a 48.5 meter building high, has an eyesight height of 1.5 meters from the top of the building, took a depression reading of the top of another nearby building and nearest wall, which are 50 degrees and 80 degrees respectively. Find the height of the nearby building in meters. The man is standing at the edge of the building and both buildings lie on the same horizontal plane. ANSWER: 39.49 meters
ring of C is 26 degrees N of E. Approximate the shortest distance of tower C to the highway. ANSWER: 374
These question were given to us as brain teaser not as homework to pass during the pandermic. The questions where surely teasing my tiny brain as I cannot get the right answer. I thought it would be easy because the answer was already given and we only need to provide the solution. I didn't even know triangles could get this tricker and confusing. BTW THANKS FOR EVERY PERSON THAT NOTICES THIS. and shout out to mark of this forum for helping me understand my trigo identities homeowrk. thanks!
Click to expand...
Could you post your diagram? What's a depression reading?
 
James Ibarra said:
1. A man standing on a 48.5 meter building high, has an eyesight height of 1.5 meters from the top of the building, took a depression reading of the top of another nearby building and nearest wall, which are 50 degrees and 80 degrees respectively. Find the height of the nearby building in meters. The man is standing at the edge of the building and both buildings lie on the same horizontal plane. ANSWER: 39.49 meters
Click to expand...
I presume "depression reading" is what I would call the "angle of depression", that is, the angle from the horizontal down to a line of sight.

But it isn't clear what "nearest wall" means. However, I get the correct answer if I take it to mean the bottom of the second building.

Sketch a picture and think about the triangle formed by the first building (a total of 50 meters from the ground), the ground, and the line of sight to the bottom of the second building. From this, you can find the distance between them.

Then try to make another triangle using the 50 degree line of sight.

Once we've dealt with this, we can talk about the next problem in your list of five. (It gets very confusing when we try to talk about several different problems in one thread.)
 
lev888 said:
Could you post your diagram? What's a depression reading?
Click to expand...
I used the given depression of 50 and 80 degrees. i put the 50 degrees into the top of the nearby building and 80 degrees to the nearest wall since i figured that the tighter the angle is based on an illustration it would have a lesser value of an angle.
 
Dr.Peterson said:
I presume "depression reading" is what I would call the "angle of depression", that is, the angle from the horizontal down to a line of sight.

But it isn't clear what "nearest wall" means. However, I get the correct answer if I take it to mean the bottom of the second building.

Sketch a picture and think about the triangle formed by the first building (a total of 50 meters from the ground), the ground, and the line of sight to the bottom of the second building. From this, you can find the distance between them.

Then try to make another triangle using the 50 degree line of sight.

Once we've dealt with this, we can talk about the next problem in your list of five. (It gets very confusing when we try to talk about several different problems in one thread.)
Click to expand...
1587930341309.png

sorry for using microsoft paint. but this is the way i drew the illustration.
Does the problem mean by "the nearest wall" the foot of the building the man is looking at?
 
It's a rather simple problem. let h = height of the nearby building, x= the distance between the buildings. Now using your diagramScreenshot 2020-04-27 at 9.59.50 AM.png

Tan 80°=50/x

x= ?

Tan 50°= (50-h)/x (solve this to get the final answer of (39.4)
 
Aliabla said:
It's a rather simple problem. let h = height of the nearby building, x= the distance between the buildings. Now using your diagramView attachment 18214

Tan 80°=50/x

x= ?

Tan 50°= (50-h)/x (solve this to get the final answer of (39.4)
Click to expand...
Hi, sorry I'm only replying now, I contracted the corona virus. I am already a negative patient but is still being quarantined for 2 weeks just to be sure, but good news is my cousin brought my laptop! so I can learn again. Sorry, I was this late in replying, I was quarantined right after i posted this problem.
 
You must log in or register to reply here.
Top