Construct a triangle ABC, in which angle B =60 degree, angle C =45 degree and AB +BC+CA=11cm
This depends on what your definition of "construct" is.
If you are construction in the classical sense, that is, you are allowed only
a straight edge and a compass this becomes a really neat question.
If you allowed some type of measurement of lengths and a protractor : ....
make a sketch of your triangle, label the angles and label the sides as
a, b, and c
let a have an arbitrary length of 1,
c/sin45 = 1/sin75
c = approx .73205
b/sin60 = 1/sin75
b = approx .8966
so the sides are in the ratio of
a : b : c = 1 : .8966 : .73205 or x : .8966x : .73205x
x + .8966x + .73205x = 11
x = 4.158
a = 4.1847
b = 3.7519
c = 3.0634
so start with the base of a = 4.1847, and construction the triangle.
Depending on the accuracy of your measurement, the angles will be forced to be 75°, 60° and 45°.
If you need a "classical" construction, ..... next post
continued ....
(sorry about the typos in the first few sentences, should have checked, hate
it when "spellcheck" takes over)
I will assume you know how to construct both a 60° and a 45° using only
a straight edge and a compass.
We will need the length of sides exactly in terms of radicals
Going back to your sketch:
we will need sin75°
= (√2 + √6)/4 , (should be part of your repertoire of popular trig ratios)
then b = sin60°/sin75° = √3/2 ÷ ( (√2+√6)/4 ) = 2√3/(√2+√6)
rationalizing and simplifying that reduces it to
b = (√18 - √6)/2
similarly: c = √3 - 1
Now the ancient Greeks had an arithmetic system where all operations such as addition, subtraction and even multiplication and division could be done by means of line segments.
Addition would be quite easy, you just attach line segments, same for subtraction.
You would start with some arbitrary length of 1 unit , call it PQ
Extend that line PQ and step off PQ with your compass so you have a line
of 2. Construction a perpendicular and step of 1 unit.
You now have a right-angled triangle with base of 2 and height of 1, giving you a hypotenuse of √3 by Pythagoras.
On that √3 hypotenuse step back (subtraction) 1 unit.
the left over part of the hypotenuse is √3-1 which is our c value.
We can now construct a triangle similar to our needed triangle.
Start with a base side of 1, at one of its ends construct a 45° angle, one arm with be the base of 1, step off the other arm from the √3-1 length you found.
and voila you have a triangle with the required angles and the correct ratio of the required sides.
One BIG problem, our perimeter is not 11.
the perimeter of our triangle is 1 + √3-1 + (√18 - √6)/2
= (2√3 + √18 - √6)/2 , but we want this to be 11
So our "scaling factor" has to be 11 ÷ ( (2√3 + √18 - √6)/2 )
= 22/(2√3 + √18 - √6)
which happens to be approx 4.1847 , the same as our side a in my first reply.
So "all we have to do" is extend our sides by that scale factor to get the perimeter to be 11.
I realize that this reply is getting out of hand and probably waaayyyy beyond
the scope of the question.
But it was fun doing it, especially since we came up with the correct ratio at the end
To construct a triangle ABC with angle B = 60 degrees, angle C = 45 degrees, and AB + BC + CA = 11 cm, follow these steps:
1. Draw a base line segment AB of any length. Label point A as the starting point and point B as the endpoint of the base.
2. From point A, draw a ray AC making an angle of 45 degrees with AB. To do this, use a protractor to measure 45 degrees and mark a point C along the ray.
3. From point B, use a protractor to draw a ray BD making an angle of 60 degrees with AB. Extend the ray beyond point D.
4. To find point P, where ray AC intersects ray BD, draw arcs with center A and radius greater than half of AB, and center B and radius greater than half of BC. The intersections of these arcs determine point P.
5. Connect points C and P to form line segment CP.
6. Measure AB, BC, and CA to find their lengths.
7. Verify that the sum of AB, BC, and CA equals 11 cm.
Now, you have successfully constructed triangle ABC with the given conditions.