MHB Drawing Triangles:Get Step-by-step Help Now!

[Redinorun]

Hello, I would need some help with drawing these 3 triangles which are part of my homework.
Would be nice if you could just help me to at least start drawing (what to do first) the right way. Thanks.

1) Regular triangle:
b - a = 3cm
c = 6cm
γ = 120°

2) Regular triangle:
a + b + c = 16cm
\alpha = 75°
β = 60°

3) Isosceles triangle:
a + h = 8cm
γ = 30° (I found out that on hand of that we know that \alpha = β = 75°)

 

[earboth]

Hello, I would need some help with drawing these 3 triangles which are part of my homework.
Would be nice if you could just help me to at least start drawing (what to do first) the right way. Thanks.

1) ...

2) Regular triangle:
a + b + c = 16cm
\alpha = 75°
β = 60°

3) ...

Hello,

these constructions are really tricky - at least for me. I can help you with 2):
View attachment 3339

1st step: Draw a line of 16 cm. Choose an arbitrary point A'. Draw an angle of 75°.

2nd step: Draw at the ends of the 16-cm-line the angles of 37.5° and 22.5°. (Do you know why that's correct?) The point of intersection of these 2 lines is the point B of your triangle.

3rd step: Draw a line through B parallel to the leg of the angle at A'. The point of intersection with the 16-cm-line is point A of your triangle.

4th step: Draw an agle of 60° at point B. The leg of this angle intersects with the 16-cm-line at point C.

5th step: Phooo!

 

[earboth]

Hello, I would need some help with drawing these 3 triangles which are part of my homework.
Would be nice if you could just help me to at least start drawing (what to do first) the right way. Thanks.

1) Regular triangle:
b - a = 3cm
c = 6cm
γ = 120°

...

Good evening,

I've drawn a sketch of the triangle:

1st step: Triangle DBC is an isosceles triangle with $$\gamma=120^\circ$$. Therefore the angles at the base must equal, that means 30°. Consequently the angle $$\angle(ADB)=150^\circ$$. Draw the triangle ABD: You know the lengthes of two sides and the angle opposite the longer side.
2nd step: Draw the base angles of $$\Delta(DBC)$$.
3rd step: The point of intersection of the leg of $$\angle[DBC)$$ with the prolonged side AD is the point C.View attachment 3341

 

[earboth]

Hello, I would need some help with drawing these 3 triangles which are part of my homework.
Would be nice if you could just help me to at least start drawing (what to do first) the right way. Thanks.

...

3) Isosceles triangle:
a + h = 8cm
γ = 30° (I found out that on hand of that we know that \alpha = β = 75°)

Good evening,

could please specify which of the three heights are meant. Two of them have equal length, but one, the angle bisector of $$\gamma$$, has (normally) a different length.

 

[earboth]

Hello, I would need some help with drawing these 3 triangles which are part of my homework.
Would be nice if you could just help me to at least start drawing (what to do first) the right way. Thanks.

...

3) Isosceles triangle:
a + h = 8cm
γ = 30° (I found out that on hand of that we know that \alpha = β = 75°)

Hello,

acoording to your answer I assume that h means the height perpendicular to AB.
View attachment 3348
1. Draw a line FT of 8 cm. (F = foot point of the height, T = top of the line)
2. Draw a line through F, perpendicular to FT.
3. Draw an angle at T with 7.5°.
4. The leg of this angle intersect the perpendicular line through F in B.
5. The perpendicular bisector of TB intersects FT in C.
6. Complete the triangle.

 

[earboth]

Hello, I would need some help with drawing these 3 triangles which are part of my homework.
Would be nice if you could just help me to at least start drawing (what to do first) the right way. Thanks.

...

3) Isosceles triangle:
a + h = 8cm
γ = 30° (I found out that on hand of that we know that \alpha = β = 75°)

Hello,

here comes the 2nd solution of question 3):
View attachment 3350

Since $$|a| = |b|$$ and $$|\alpha| = |\beta|$$ I've used the side b to finde the point A.
The numbers indicate the steps of the construction.

 

[johng1]

Hi,
Earboth has given some ingenious constructions. However, assuming that any constructible length is available, your drawing problems all have straight forward solutions.

Ruler and Compass Construction

I assume that a line segment of unit length 1 can be constructed. A positive number a is constructible if and only if a line segment of length a can be constructed with ruler and compass. If a and b are positive constructible numbers, a+b, ab and a/b are constructible. Also $\sqrt{a}$ is constructible; also if a>b, a-b is constructible. Here is a link which shows the elementary geometric constructions as well as other interesting facts. Reduction: Constructible Numbers Certain angles are constuctible. For example, angles of 30 degrees, 45 degrees, 60 degrees are constructible. If angles $\alpha$ and $\beta$ are constructible, so also are $\alpha\pm\beta$. If an angle $\alpha<90$ is constructible, then $\cos(\alpha)$ and $\sin(\alpha)$ are constructible numbers -- draw a circle with diameter 1. Then construct angle $\alpha$ at one end of the diameter and intersect the other side of the angle with the circle. Clearly, $\cos(\alpha)$ and $\sin(\alpha)$ are the lengths of the sides of the right triangle.