# Thread: How to construct triangle

1. ## How to construct triangle

$c - a = 2cm$
$b = 5 cm$
$\gamma = 30^0$

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

2. You probably should:

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

Are a, b, and c the three sides?

Are $\alpha$, $\beta$, and $\gamma$ 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?

3. sorry about that.
yeah a,b,c are the sides.

is opposite of a
is opposite of b
is opposite of c

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

+ + = 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.

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

is opposite of a
is opposite of b
is opposite of c

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

+ + = 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 your base, b=5
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).

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

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

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

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

simplify these to find "a".

If you multiply out, you get

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

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

$25-10x=4a+4$

$21-10x=4a$

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

$21-10acos30^o=4a$

solve for "a" which then gives "c"