# finding largest value of tan angle

• Dec 18th 2009, 09:56 AM
Vicky1997
finding largest value of tan angle
Triangle ABC has angle C = 60 degrees and BC = 4. Point D is the midpoint of BC. What is the largest possible value of tan ( angle BAD ) ?

I have been working on this problem for over a week and I understand it is way over my level, but I can't give up. It's frustrating because the problem seems to be so simple. ( but now i see it looked simple because of my lack of knowledge)

Hoping to get some ideas, I have been studying Trigonometry and Geometry( triangle section only).

So far, I was looking for a way to make DA as small as I can.
Wouldn't that make tan (angle BAD ) the largest?
Or am I hopelessly off?

• Dec 18th 2009, 10:53 AM
Raoh
Quote:

Originally Posted by Vicky1997
Triangle ABC has angle C = 60 degrees and BC = 4. Point D is the midpoint of BC. What is the largest possible value of tan ( angle BAD ) ?

I have been working on this problem for over a week and I understand it is way over my level, but I can't give up. It's frustrating because the problem seems to be so simple. ( but now i see it looked simple because of my lack of knowledge)

Hoping to get some ideas, I have been studying Trigonometry and Geometry( triangle section only).

So far, I was looking for a way to make DA as small as I can.
Wouldn't that make tan (angle BAD ) the largest?
Or am I hopelessly off?

hi(Happy)
you have $\tan(\widehat{BAD})=\frac{BD}{AD}$
what is the largest value of $\frac{BD}{AD} ?$
• Dec 18th 2009, 05:00 PM
Vicky1997
Quote:

Originally Posted by Raoh
hi(Happy)
you have $\tan(\widehat{BAD})=\frac{BD}{AD}$
what is the largest value of $\frac{BD}{AD} ?$

I am not sure how to get it.
Could you please explain to me how to find the largest value?

Vicky.
• Dec 18th 2009, 06:46 PM
Soroban
Hello, Vicky!

I hope I understand the problem . . .

Quote:

Triangle $ABC$ has: . $\angle C = 60^o,\;BC = 4.$
Point $D$ is the midpoint of $BC.$
What is the largest possible value of $\tan(\angle BAD)$ ?

Code:

                C                 o               *  *               * 60° * 2             *        *             *          *  D           *              o           *          *  | *         *        *      |  * 2         *    *          |    *       *  *  θ            |      *     A o - - - - - - - - - o - - - - o B                           E
We have: . $\angle C = 60^o,\;CD = DB = 2$
Let $\theta = \angle BAD.$
Draw $DE \perp AB.$

And we want to maximize: . $\tan\theta \,=\,\frac{DE}{AE}$

Is this correct?

• Dec 18th 2009, 07:53 PM
Vicky1997
Quote:

Originally Posted by Soroban
Hello, Vicky!

I hope I understand the problem . . .

Code:

                C                 o               *  *               * 60° * 2             *        *             *          *  D           *              o           *          *  | *         *        *      |  * 2         *    *          |    *       *  *  θ            |      *     A o - - - - - - - - - o - - - - o B                           E
We have: . $\angle C = 60^o,\;CD = DB = 2$
Let $\theta = \angle BAD.$
Draw $DE \perp AB.$

And we want to maximize: . $\tan\theta \,=\,\frac{DE}{AE}$

Is this correct?

Thanks to you, I think I am starting to understand the problem better.
But all I have is the answer.
Would your solution get root 3 over 4 root 2 minus 3 ?

Vicky.
• Dec 19th 2009, 02:26 AM
Hello Vicky1997

This is quite tricky - I've been puzzling over it for some time. But I have a solution - and it agrees with the one you've been given.

In the attached diagram, I have called the length of $AC, x$ cm, and I've drawn the perpendiculars from $C$ and $M$ (the mid-point of $BC$) to $AB$, meeting it at $E$ and $D$.

Here's my solution.

The Cosine Rule on $\triangle ABC$ gives:
$AB^2 = x^2-4x+16$ (1)
The area of $\triangle ABC$ can be written in two ways, using the formulae $\tfrac12AC.BC\sin \angle ACB$ and $\tfrac12AB.EC$. This gives:
$\tfrac124x\sin60^o=\sqrt3x=\tfrac12AB.EC$

$\Rightarrow EC = \frac{2\sqrt3x}{AB}$

$\Rightarrow DM = \frac{\sqrt3x}{AB}$, using the similar triangles $BMD, BCE$ (2)
Next, using Pythagoras on $\triangle BMD$:
$DB^2=MB^2-DM^2$
$= 4 - \frac{3x^2}{AB^2}$, from (2)

$= \frac{4AB^2-3x^2}{AB^2}$

$=\frac{x^2-16x+64}{AB^2}$, from (1)

$=\frac{(8-x)^2}{AB^2}$
$\Rightarrow DB = \frac{8-x}{AB}$ (3)
You'll see that I've taken this sign, rather than $(x-8)$, since $x>8$ will clearly not maximise the value of $\tan\theta$.

Now:
$AD = AB-DB$
$= \frac{AB^2-(8-x)}{AB}$, from (3)

$=\frac{x^2-3x+8}{AB}$, from (1)
and:
$\tan\theta = \frac{MD}{AD}$
$= \frac{\sqrt3x}{x^2-3x+8}$ (4)
All the hard work is done now. It's a simple matter to differentiate this function, and equate the result to zero for a maximum or minimum. When you do this, the result simplifies to:
$-x^2+8=0$

$\Rightarrow x = 2\sqrt2$
You can check that this gives a maximum value of $\tan\theta$, and that value, from (4), is:
$\frac{2\sqrt2\sqrt3}{8-6\sqrt2+8}$
$=\frac{\sqrt3}{4\sqrt2-3}$
• Dec 19th 2009, 09:03 AM
Vicky1997
Quote:

Hello Vicky1997

This is quite tricky - I've been puzzling over it for some time. But I have a solution - and it agrees with the one you've been given.

In the attached diagram, I have called the length of $AC, x$ cm, and I've drawn the perpendiculars from $C$ and $M$ (the mid-point of $BC$) to $AB$, meeting it at $E$ and $D$.

Here's my solution.

The Cosine Rule on $\triangle ABC$ gives:
$AB^2 = x^2-4x+16$ (1)
The area of $\triangle ABC$ can be written in two ways, using the formulae $\tfrac12AC.BC\sin \angle ACB$ and $\tfrac12AB.EC$. This gives:
$\tfrac124x\sin60^o=\sqrt3x=\tfrac12AB.EC$

$\Rightarrow EC = \frac{2\sqrt3x}{AB}$

$\Rightarrow DM = \frac{\sqrt3x}{AB}$, using the similar triangles $BMD, BCE$ (2)
Next, using Pythagoras on $\triangle BMD$:
$DB^2=MB^2-DM^2$
$= 4 - \frac{3x^2}{AB^2}$, from (2)

$= \frac{4AB^2-3x^2}{AB^2}$

$=\frac{x^2-16x+64}{AB^2}$, from (1)

$=\frac{(8-x)^2}{AB^2}$
$\Rightarrow DB = \frac{8-x}{AB}$ (3)
You'll see that I've taken this sign, rather than $(x-8)$, since $x>8$ will clearly not maximise the value of $\tan\theta$.

Now:
$AD = AB-DB$
$= \frac{AB^2-(8-x)}{AB}$, from (3)

$=\frac{x^2-3x+8}{AB}$, from (1)
and:
$\tan\theta = \frac{MD}{AD}$
$= \frac{\sqrt3x}{x^2-3x+8}$ (4)
All the hard work is done now. It's a simple matter to differentiate this function, and equate the result to zero for a maximum or minimum. When you do this, the result simplifies to:
$-x^2+8=0$

$\Rightarrow x = 2\sqrt2$
You can check that this gives a maximum value of $\tan\theta$, and that value, from (4), is:
$\frac{2\sqrt2\sqrt3}{8-6\sqrt2+8}$
$=\frac{\sqrt3}{4\sqrt2-3}$

Thank you!!!!(Rofl)

Honestly, I only understand certain parts of the solution. But I have printed it out and this is what I will be studying over the winter break.

Vicky.