# How to construct triangle

• Apr 22nd 2010, 12:26 PM
metlx
How to construct triangle
$\displaystyle c - a = 2cm$
$\displaystyle b = 5 cm$
$\displaystyle \gamma = 30^0$

I know how to solve if you have a-b = x.. you construct a isosceles triangle and then continue from there.. but when you have a hypotenuse it's different..

can someone help?
dot steps would be great :)
• Apr 22nd 2010, 01:21 PM
TKHunny
You probably should:

1) Draw the triangle.
2) Label the parts.

Are a, b, and c the three sides?

Are $\displaystyle \alpha$, $\displaystyle \beta$, and $\displaystyle \gamma$ the angles opposite a, b, and c respectively?

Get in the habit of defining what it is you are doing. Not everything has a "standard" definition.

Do you have the Law of Sines, Cosines?
• Apr 22nd 2010, 02:46 PM
metlx
yeah a,b,c are the sides.

http://www.mathhelpforum.com/math-he...49852f08-1.gif is opposite of a
http://www.mathhelpforum.com/math-he...ed813421-1.gif is opposite of b
http://www.mathhelpforum.com/math-he...65c1de79-1.gif is opposite of c

no law of sines nor cosine.
only use of basic things:

http://www.mathhelpforum.com/math-he...49852f08-1.gif + http://www.mathhelpforum.com/math-he...ed813421-1.gif + http://www.mathhelpforum.com/math-he...65c1de79-1.gif = 180º

a + b > c
(a and b are catheti and c is hypotenuse)
+ all the facts about various triangles. (right, oblique,equilateral, isosceles, scalene)

the construction is supposed to be made without "calculating" anything, just the process of how to draw a triangle with the given info.
(no need to draw it, just need someone to explain the steps)..

what i'd do first is:

1) draw the b (=5cm)
2) measure gamma (=30º)
3) draw the line on which a will be.
4) ... < stuck here.
• Apr 22nd 2010, 03:10 PM
Quote:

Originally Posted by metlx
yeah a,b,c are the sides.

http://www.mathhelpforum.com/math-he...49852f08-1.gif is opposite of a
http://www.mathhelpforum.com/math-he...ed813421-1.gif is opposite of b
http://www.mathhelpforum.com/math-he...65c1de79-1.gif is opposite of c

no law of sines nor cosine.
only use of basic things:

http://www.mathhelpforum.com/math-he...49852f08-1.gif + http://www.mathhelpforum.com/math-he...ed813421-1.gif + http://www.mathhelpforum.com/math-he...65c1de79-1.gif = 180º
[a2 + b2 = c2]
a + b > c
(a and b are catheti and c is hypotenuse)
+ all the facts about various triangles. (right, oblique,equilateral, isosceles, scalene)

the construction is supposed to be made without "calculating" anything, just the process of how to draw a triangle with the given info.
(no need to draw it, just need someone to explain the steps)..

what i'd do first is:

1) draw the b (=5cm)
2) measure gamma (=30º)
3) draw the line on which a will be.
4) ... < stuck here.

Hi metlx,

Draw the side "a" on the left at an angle of 30 degrees.

Since c-a=2, then c=a+2, so c > a.

Hence draw the 3rd side a little longer than "a" touching that line.

The drawing is just approximate to begin with.

Now draw a verical line from the top of the triangle to the base "b" and call it "h".

Where that touches the base, label the part of the base to the left "x"
and the part to the right "5-x".

Now you have 2 back-to-back right-angled triangles.
You can apply Pythagoras' theorem and Sin(30), cos(30), tan(30).

$\displaystyle sin30^o=\frac{h}{a}$

$\displaystyle cos30^o=\frac{x}{a}$

$\displaystyle (5-x)^2+h^2=(a+2)^2$

$\displaystyle x^2+h^2=a^2$

simplify these to find "a".

If you multiply out, you get

$\displaystyle 25-10x+x^2+h^2=a^2+4a+4$

$\displaystyle 25-10x+a^2=a^2+4a+4$

$\displaystyle 25-10x=4a+4$

$\displaystyle 21-10x=4a$

$\displaystyle cos30^o=\frac{x}{a},\ x=acos30^o$

$\displaystyle 21-10acos30^o=4a$

solve for "a" which then gives "c"