- #1

DrLiangMath

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AB=3, DC=5,

**∠**CAD=$45^o$, AB ⊥ BC. Find the length of AC.You are using an out of date browser. It may not display this or other websites correctly.

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- MHB
- Thread starter DrLiangMath
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- #1

DrLiangMath

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AB=3, DC=5, **∠** CAD=$45^o$, AB ⊥ BC. Find the length of AC.

- #2

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It's inefficient but start by defining angle BCA to be C. Then angle CDA = 135 - C. Call x = AC. Then by Law of Sines

\(\displaystyle \dfrac{sin(90)}{x} = \dfrac{sin(45)}{5}\)

\(\displaystyle \dfrac{sin(135 - C)}{x} = \dfrac{sin(45)}{5}\)

This gives a biquadratic in x. But, as I said, it's inefficient. There is probably a better approach.

-Dan

- #3

DrLiangMath

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Thank you for your response. But we still need one more equation since we have 2 variables x and C.

- #4

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I gave two equations. The first is the Law of Sines in triangle ABC and the second is the Law of Sines in triangle ADC. Perhaps I should have stated that.Thank you for your response. But we still need one more equation since we have 2 variables x and C.

-Dan

- #5

I like Serena

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MHB

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Are you actually looking for an answer to this problem? Or do you have an elegant answer to share?

If the latter, then we might move this thread to the https://mathhelpboards.com/forums/challenge-questions-and-puzzles.28/ subforum.

Btw, at this time I could provide an answer, but it's rather long winded and I'm still looking for a more elegant solution.

- #6

I like Serena

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MHB

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Here's the best I could find.

\begin{tikzpicture}

%preamble \usetikzlibrary {angles,quotes}

\coordinate[label=left:A] (A) at (0,3);

\coordinate[label=left:B] (B) at (0,0);

\coordinate[label=right:C] (C) at (6,0);

\coordinate[label=below: D] (D) at (1,0);

\draw (B) rectangle +(0.2,0.2);

\draw[thick] (A) -- (B) -- (C) -- cycle (A) -- (D);

\path (A) -- node[ left ] {3} (B) -- node[below] {$y$} (D) -- node[below] {5} (C) -- node[above right] {$x$} (A);

\pic ["$45^\circ$", draw, angle radius=0.4cm, angle eccentricity=2.2,pic text options={shift={(0.1,0)}}] {angle = D--A--C};

\pic ["$135^\circ-C$", draw, angle radius=0.3cm, angle eccentricity=2.2,pic text options={shift={(0.5,0)}}] {angle = C--D--A};

\end{tikzpicture}

Let $x=AC$, which is what we want to find.

Let $y=BD$.

From the law of sines we have:

$$\frac 5{\sin 45}=\frac x{\sin(135-C)}

\implies x=5\cdot\frac{\sin(135-C)}{\sin 45}=5\cdot\frac{\sin 135\cos C -\cos 135\sin C}{\sin 45}=5\cdot\frac{\sin 45\cos C +\cos 45\sin C}{\sin 45}=5(\cos C+\sin C) \tag 1$$

From the definitions of cosine and sine:

$$\cos C=\frac{BD+5}{x}=\frac{y+5}{x} \tag 2$$

$$\sin C=\frac 3{x} \tag 3$$

From Pythagoras:

$$x^2 = (y + 5)^2 + 3^2 \tag 4$$

Substitute (2) and (3) into (1) to find:

$$x=5\left(\frac{y+5}x+\frac 3x\right) \implies x^2=5(y+8)\tag 5$$

Substitute in (4):

$$5(y+8) = (y + 5)^2 + 3^2 \implies y^2+5y-6= (y-1)(y+6) = 0 \implies y =1 \tag 6$$

Substitute back into (5):

$$x^2=5(1+8)=45 \implies x=3\sqrt 5$$

which is the answer.

\begin{tikzpicture}

%preamble \usetikzlibrary {angles,quotes}

\coordinate[label=left:A] (A) at (0,3);

\coordinate[label=left:B] (B) at (0,0);

\coordinate[label=right:C] (C) at (6,0);

\coordinate[label=below: D] (D) at (1,0);

\draw (B) rectangle +(0.2,0.2);

\draw[thick] (A) -- (B) -- (C) -- cycle (A) -- (D);

\path (A) -- node[ left ] {3} (B) -- node[below] {$y$} (D) -- node[below] {5} (C) -- node[above right] {$x$} (A);

\pic ["$45^\circ$", draw, angle radius=0.4cm, angle eccentricity=2.2,pic text options={shift={(0.1,0)}}] {angle = D--A--C};

\pic ["$135^\circ-C$", draw, angle radius=0.3cm, angle eccentricity=2.2,pic text options={shift={(0.5,0)}}] {angle = C--D--A};

\end{tikzpicture}

Let $y=BD$.

From the law of sines we have:

$$\frac 5{\sin 45}=\frac x{\sin(135-C)}

\implies x=5\cdot\frac{\sin(135-C)}{\sin 45}=5\cdot\frac{\sin 135\cos C -\cos 135\sin C}{\sin 45}=5\cdot\frac{\sin 45\cos C +\cos 45\sin C}{\sin 45}=5(\cos C+\sin C) \tag 1$$

From the definitions of cosine and sine:

$$\cos C=\frac{BD+5}{x}=\frac{y+5}{x} \tag 2$$

$$\sin C=\frac 3{x} \tag 3$$

From Pythagoras:

$$x^2 = (y + 5)^2 + 3^2 \tag 4$$

Substitute (2) and (3) into (1) to find:

$$x=5\left(\frac{y+5}x+\frac 3x\right) \implies x^2=5(y+8)\tag 5$$

Substitute in (4):

$$5(y+8) = (y + 5)^2 + 3^2 \implies y^2+5y-6= (y-1)(y+6) = 0 \implies y =1 \tag 6$$

Substitute back into (5):

$$x^2=5(1+8)=45 \implies x=3\sqrt 5$$

which is the answer.

Last edited:

- #7

DrLiangMath

- 22

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Are you actually looking for an answer to this problem? Or do you have an elegant answer to share?

If the latter, then we might move this thread to the https://mathhelpboards.com/forums/challenge-questions-and-puzzles.28/ subforum.

Btw, at this time I could provide an answer, but it's rather long winded and I'm still looking for a more elegant solution.

I do have a solution to share. Different form the solution given by

- #8

I like Serena

Homework Helper

MHB

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TikZ is natively supported on this site similar to formulas. Click on the picture to see the $\LaTeX$ code.

\begin{tikzpicture}[scale=2]

%preamble \usepackage{tkz-euclide}

\tkzDefPoint(0,3){A}

\tkzDefPoint(0,0){B}

\tkzDefPoint(6,0){C}

\tkzDefPoint(1,0){D}

\tkzDrawSegments[ultra thick](A,B B,C C,A A,D)

\tkzDefPointWith[orthogonal,K=-1](D,A)

\tkzGetPoint{E}

\tkzDefPointBy[projection=onto B--C](E)

\tkzGetPoint{F}

\tkzDrawSegments[help lines](D,E E,F)

\tkzLabelPoints[ left ](A,B)

\tkzLabelPoints[ right ](C)

\tkzLabelPoints[ below ](D)

\tkzLabelPoints[ below,help lines ](F)

\tkzLabelPoints[ above right,help lines ](E)

\tkzLabelSegment(B,A){3}

\tkzLabelSegment(C,D){5}

\tkzLabelSegment(D,B){$x$}

\tkzLabelSegment[help lines](E,F){$x$}

\tkzLabelSegment[help lines](D,F){3}

\tkzLabelSegment[help lines](F,C){2}

\tkzLabelAngles[pos=1](D,A,C){$45^\circ$}

\tkzMarkRightAngles(A,B,C)

\tkzMarkAngles[size=0.7,mark=none](D,A,C)

\tkzLabelAngles[pos=1,help lines](A,E,D){$45^\circ$}

\tkzMarkAngles[size=0.7,mark=none,help lines](A,E,D)

\tkzMarkAngles[size=0.6,mark=||,help lines](B,A,D F,D,E)

\tkzMarkRightAngles[help lines](A,D,E E,F,D)

\tkzMarkSegments[mark=|,help lines](A,D D,E)

\end{tikzpicture}

- #9

DrLiangMath

- 22

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Wow, it looks so nice! Thank you!

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