# Geometry & Trigonometry - Two solutions

1. Aug 3, 2015

### Rectifier

The problem
a) find the length of AD in the figure
b) Why are there two solutions in a) and what solution fits the figure?
Figure

The attempt

I started with drawing a "help line" in the figure.

The cosine formula for AC with respect to triangles ADC and CBA gives us two equations:

$AC^2 = 4.1^2 + 3.7^2 - 2 \cdot 3.7 \cdot 4.1 \cdot cos(57)$
and
$AC^2 = 4.3^2 + x^2 - 2 \cdot 4.3 \cdot x \cdot cos(60)$

Setting these two equal to each other gives a polynomial of the 2nd degree that has two roots $x_1=2.47886$ and $x_2=1.82114$. (These solutions are correct)

I am stuck at b). I have no idea how to continue.

Last edited: Aug 3, 2015
2. Aug 3, 2015

### SammyS

Staff Emeritus
What are all the angles that result from each solution?

Is the given drawing close to being drawn to scale?

3. Aug 3, 2015

### Rectifier

No, the drawing is not drawn to scale. Should I calculate the all the possible angles in the figure?

4. Aug 3, 2015

### SammyS

Staff Emeritus
That could help, especially for the portion above the "help line".

5. Aug 3, 2015

### Rectifier

Skip to the next comment. It seems like this one is comepletely wrong.

Here comes my failed attempt:

Attempt:
I declare two angles above the "help line"
$∠A_1$ and $∠C_1$

$\frac{sin(60)}{AC}= \frac{sine(A_1)}{4.3} = \frac{sin(C_1)}{x}$

$AC^2=13.9757 \\ AC = 3.73841$

I know that $A_1, C_1 < 120$ $\ \ \ \ (180-60)$ since $∠D = 60$
$∠A_1$ is either $15.4535 \ or \ 164.547$
$∠A_1=164$ is outside of the interval
$∠C_1$ was harder to get since there are two x:es

$x_1=2.47886$
$x_1\frac{sin(60)}{AC}=sin(C_1) \\ C_1a = 6.47961 \\ C_1aa = 173. 52$
$C_1aa$ is outside the interval above

$x_2=1.82114$
$x_2\frac{sin(60)}{AC}=sin(C_1) \\ C_1b = 8.83598 \\ C_1bb = 171. 164$
$C_1bb$ is outside the interval above

Both of these angels are wrong since both of them $+ 60 \neq 180$

Further attempt:
The main angles $∠A$ and $∠C$ are
$∠A + ∠C = 360 - 60 - 57 = 243$

Last edited: Aug 3, 2015
6. Aug 3, 2015

### Rectifier

It seemes like my calculations were wrong above.

Attempt 2:

$180 = A_1 + C_1 + 60 \\ 120 = A_1 + C_2 \ \ A_1, C_1 < 120$
$A_1 = sin ^{-1} 4.3 \frac{sin(60)}{AC}=84.9519 \\ and \\ A_1 = 180 - 84.9519 = 95.0481$

thus
$A_1 = 84.9519 \ or \ 95.0481$

$C_1 = 120 - A_1$

ONE:
$C_1 = 120 - 84.9519 = 35.0481$

TWO:
$C_1 = 120 - 95.0481 = 24.9512$

7. Aug 3, 2015

### SammyS

Staff Emeritus
Somehow you've made a mistake in finding angle A1.

Did you use AC2 rather than AC by mistake?

You will find two reasonable values for angle A1. Each gives a values for C. Do those agree with the two x values?

You may want to find A2 and C2 (below the help line).

Added in Edit: I just saw your recent post.

8. Aug 3, 2015

9. Aug 3, 2015

### SammyS

Staff Emeritus
Just to check yourself. Do those agree with the two x values?

10. Aug 3, 2015

### Rectifier

Yah, I checked both of them and got the x:es that I got before.

11. Aug 3, 2015

### SammyS

Staff Emeritus
You may want to find A2 and C2 (below the help line). Then find the overall angles, A and C .

So far there doesn't seem to be any reason to eliminate either of the two values for x, at least, not that I can see.

12. Aug 3, 2015

### insightful

This may be obvious to you, but if you draw a circle centered at C with radius AC, the source of the two solutions and the answer to b) become clear.

13. Aug 4, 2015

### Rectifier

What do you mean by "the source of two solutions"? I am bad at english :,(

Last edited: Aug 4, 2015
14. Aug 4, 2015

### insightful

I mean it graphically shows the reason why there are two solutions, and why one of them matches the original sketch and the other one doesn't.

15. Aug 4, 2015

### Rectifier

Is this drawing correct?

Not sure how the corners D and B should be placed on the circle.

I am not sure how to see that there are two solutions now... :,(

16. Aug 4, 2015

### insightful

B is not relevant.

Yeah, the drawing is so out of scale, it makes it hard. If it were to scale (I suggest you redraw it), the circle will cross AD much closer to A and then moving A to this new point is the second answer. The first answer has angle DAC as an acute angle and matches the original sketch. The second answer has angle DAC as an obtuse angle and so does not match.

17. Aug 4, 2015

### andrewkirk

we can take the points A, B C to be fixed, as triangle ABC is fully specified.

Consider what happens to the angle $\angle ADC$ if we allow it to vary (rather than fixing it at 60 deg) as angle $\angle ACD$ varies from 0 to 180 degrees, keeping the length of CD fixed at 4.3. CD is longer than AC so at $\angle ACD=0$ D will be on the extension of AC at a point further from C than A. Call that point D0. So $\angle ADC=0$.

When $\angle ACD=$ 180 degrees D will be on the extension of AC at a point further from A than C. Call that point D180. So $\angle ADC=0$ again.

$\angle ADC=0$ will be at a maximum when triangle ADC is isosceles, which is when the projection of D onto AC is at the midpoint of AC. And that maximum will be more than 60 degrees. Call the place where D is in that case Disos.

So the angle must pass through 60 degrees twice, once as D rotates from D0 towards Disos, and once again as D rotates from Disos towards D180. Those two points are the two solutions.

From the diagram, which I haven't checked for scale, it looks like the solution drawn is the first of the two solutions. IN the other solution $\angle ACD$ will be of the order of 140 degrees.

18. Aug 5, 2015

### insightful

For better or worse, here are my two solutions, drawn to scale. You can faintly see an arc of the circle through A I referred to above.

#### Attached Files:

• ###### Polygons.jpg
File size:
28.9 KB
Views:
71
19. Aug 6, 2015

### DeldotB

Well you could: draw a line from D to B and use Pythagorean theorem to find length DB. Then use law of cosines to solve for length DA....( assuming that's a right angle there)

20. Aug 6, 2015

### SammyS

Staff Emeritus
Where is there a right triangle here ?