Mixing
Finding Coordinates of a Point That Creates a Right Angle
$begingroup$
The question is as follows:
Let $A$ = $(0, 0)$ and $C$ = $(4, 3)$. Point $D$ is located so that angle $ACD$ is a right angle and the tangent of angle $DAC$ is 3/4. Find coordinates for $D$. There are two answers.
I was just able to find the slope of AC, which is 3/4. So that means that the slope perpendicular to that would be $-frac{4}{3}$. But I don't know how to go any further than that and I also am unsure about how to use the tangent of angle $DAC$ for this problem. Any help will be greatly appreciated.
geometry trigonometry
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Let $A$ = $(0, 0)$ and $C$ = $(4, 3)$. Point $D$ is located so that angle $ACD$ is a right angle and the tangent of angle $DAC$ is 3/4. Find coordinates for $D$. There are two answers.
I was just able to find the slope of AC, which is 3/4. So that means that the slope perpendicular to that would be $-frac{4}{3}$. But I don't know how to go any further than that and I also am unsure about how to use the tangent of angle $DAC$ for this problem. Any help will be greatly appreciated.
geometry trigonometry
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
$endgroup$
$begingroup$
the slope of AC is not 5
$endgroup$
– Nosrati
Sep 15 '17 at 1:35
$begingroup$
@MyGlasses Ah yes, I found the distance of AC, not the slope! But I still do not know what to do with the tangent part of the problem.
$endgroup$
– geo_freak
Sep 15 '17 at 1:41
add a comment |
$begingroup$
The question is as follows:
Let $A$ = $(0, 0)$ and $C$ = $(4, 3)$. Point $D$ is located so that angle $ACD$ is a right angle and the tangent of angle $DAC$ is 3/4. Find coordinates for $D$. There are two answers.
I was just able to find the slope of AC, which is 3/4. So that means that the slope perpendicular to that would be $-frac{4}{3}$. But I don't know how to go any further than that and I also am unsure about how to use the tangent of angle $DAC$ for this problem. Any help will be greatly appreciated.
geometry trigonometry
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
$endgroup$
The question is as follows:
Let $A$ = $(0, 0)$ and $C$ = $(4, 3)$. Point $D$ is located so that angle $ACD$ is a right angle and the tangent of angle $DAC$ is 3/4. Find coordinates for $D$. There are two answers.
I was just able to find the slope of AC, which is 3/4. So that means that the slope perpendicular to that would be $-frac{4}{3}$. But I don't know how to go any further than that and I also am unsure about how to use the tangent of angle $DAC$ for this problem. Any help will be greatly appreciated.
geometry trigonometry
geometry trigonometry
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
edited Sep 15 '17 at 1:42
geo_freak
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
asked Sep 15 '17 at 1:33
geo_freakgeo_freak
422210
422210
$begingroup$
the slope of AC is not 5
$endgroup$
– Nosrati
Sep 15 '17 at 1:35
$begingroup$
@MyGlasses Ah yes, I found the distance of AC, not the slope! But I still do not know what to do with the tangent part of the problem.
$endgroup$
– geo_freak
Sep 15 '17 at 1:41
add a comment |
$begingroup$
the slope of AC is not 5
$endgroup$
– Nosrati
Sep 15 '17 at 1:35
$begingroup$
@MyGlasses Ah yes, I found the distance of AC, not the slope! But I still do not know what to do with the tangent part of the problem.
$endgroup$
– geo_freak
Sep 15 '17 at 1:41
$begingroup$
the slope of AC is not 5
$endgroup$
– Nosrati
Sep 15 '17 at 1:35
$begingroup$
the slope of AC is not 5
$endgroup$
– Nosrati
Sep 15 '17 at 1:35
$begingroup$
@MyGlasses Ah yes, I found the distance of AC, not the slope! But I still do not know what to do with the tangent part of the problem.
$endgroup$
– geo_freak
Sep 15 '17 at 1:41
$begingroup$
@MyGlasses Ah yes, I found the distance of AC, not the slope! But I still do not know what to do with the tangent part of the problem.
$endgroup$
– geo_freak
Sep 15 '17 at 1:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The slope of $AC$ has been calculated wrong : it is in fact $frac{3-0}{4-0} = frac 34$. Hence, the slope of the line perpendicular to $AC$ must be $frac{-4}{3}$. So any point $D$ is located on the straight line passing through $C$ and having slope $frac{-4}{3}$ i.e. $(y-3) = -frac 43(x-4)$. This simplifies to $3y+4x = 25$.
Hence, $D$ satisfies these coordinates. Furthermore, the length of $AC$ is $5$, as you have said above.
Try to imagine the triangle ACD in your mind. This triangle is right angled at $C$, and the length of $AC$ is $5$.
EDIT : What is the definition of tangent of $DAC$? It is defined as the ratio between the opposite and adjacent sides, right? That is, $tan DAC = frac{OPP}{ADJ}$, where $OPP$ is the side opposite the angle $DAC$, and $ADJ$ is the side adjacent to $DAC$ which is not the hypotenuse (for that, we use the phrase $HYP$, so this side is not the hypotenuse).
Picturing the triangle in your mind, you must be able to see that $OPP = DC$, and $ADJ = AC$. Hence, $tan DAC = frac{DC}{AC} = frac 34$. Therefore, $DC = frac{3AC}{4} = frac{3 times 5}{4} = frac{15}{4} = 3.75$.
Now, you have a right angled triangle, hence by Pythagoras' theorem, $AD^2 = AC^2+DC^2$ (from your diagram, you must have seen that $AD$ is opposite angle $C$ which is right angled, so it is the hypotenuse). From this, $AD^2 = 25 + 14.0625 = 39.0625$, hence $AD = 6.25$.
So, if $D= (x,y)$, then $x^2+y^2 = 39.0625$ by the distance from $0$ being $0.625$, and $3x+4y = 25$. Solving, you get two points : $(x,y) = (0,6.25)$, and $(x,y) = (6,1.75)$. These are the two candidate points for $D$.
POST-EDIT: If now you have understood, then I am willing to clarify two things. One, you might be wondering if the calculation to find $AD = 6.25$ is difficult. After all, we were (doing the equivalent of) squaring $3$ digit numbers there, and some six-digit arithmetic. But in fact, that part is really easy, and I can tell you why.
Secondly, how did I solve the equations $x^2+y^2 = 39.0625$ and $3x+4y=25$? That too was simple,and I can explain that as well if you have not understood yet.
POST-POST-EDIT : To solve this equation, we eliminate $y$ via $4y = 25-3x$. Multiplying the first equation by sixteen gives $16x^2+16y^2 = 625$, so $16x^2+(25-3x)^2 = 625$, a quadratic equation in $x$. From here it should be easy.
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
$begingroup$
I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
|
show 2 more comments
$begingroup$
Hint:)
Since slope of $AC$ is $dfrac34$ and $tan DAC=dfrac34$, use $tan(x+y)=dfrac{tan x+tan y}{1-tan xtan y}$ for finding the slope of $AD$. Find the intersection of lines $AD$ and $DC$.
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
You should sketch it:
$hspace{3cm}$
Using the distance formula, you can find:
$$AC=sqrt{(4-0)^2+(3-0)^2}=5 text{(which is basically Pythagoras formula)}.$$
Given $tan angle D_1AC=tan angle D_2AC=frac34$, you can find:
$$tan angle D_1AC=frac{CD_1}{AC}=frac34 Rightarrow CD_1=frac{3AC}{4}=frac{15}4=CD_2.$$
Note that the triangles $ABC$ and $BCE$ are similar, hence:
$$frac{CE}{AC}=frac{BC}{AB} Rightarrow frac{CE}{5}=frac{3}{4} Rightarrow CE=frac{15}{4}=CD_2 Rightarrow E=D_2;\
frac{BE}{BC}=frac{BC}{AB} Rightarrow BE=frac{BC^2}{AB}=frac{9}{4} Rightarrow D_2left(frac{25}{4},0right)$$
Now you can use the similarity of the triangles $BCE$ and $FD_1E$ to find the coordinates of $D_1$. It is an exercise for you. Answer is:
$D_1left(frac74,6right)$
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
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active
oldest
votes
$begingroup$
The slope of $AC$ has been calculated wrong : it is in fact $frac{3-0}{4-0} = frac 34$. Hence, the slope of the line perpendicular to $AC$ must be $frac{-4}{3}$. So any point $D$ is located on the straight line passing through $C$ and having slope $frac{-4}{3}$ i.e. $(y-3) = -frac 43(x-4)$. This simplifies to $3y+4x = 25$.
Hence, $D$ satisfies these coordinates. Furthermore, the length of $AC$ is $5$, as you have said above.
Try to imagine the triangle ACD in your mind. This triangle is right angled at $C$, and the length of $AC$ is $5$.
EDIT : What is the definition of tangent of $DAC$? It is defined as the ratio between the opposite and adjacent sides, right? That is, $tan DAC = frac{OPP}{ADJ}$, where $OPP$ is the side opposite the angle $DAC$, and $ADJ$ is the side adjacent to $DAC$ which is not the hypotenuse (for that, we use the phrase $HYP$, so this side is not the hypotenuse).
Picturing the triangle in your mind, you must be able to see that $OPP = DC$, and $ADJ = AC$. Hence, $tan DAC = frac{DC}{AC} = frac 34$. Therefore, $DC = frac{3AC}{4} = frac{3 times 5}{4} = frac{15}{4} = 3.75$.
Now, you have a right angled triangle, hence by Pythagoras' theorem, $AD^2 = AC^2+DC^2$ (from your diagram, you must have seen that $AD$ is opposite angle $C$ which is right angled, so it is the hypotenuse). From this, $AD^2 = 25 + 14.0625 = 39.0625$, hence $AD = 6.25$.
So, if $D= (x,y)$, then $x^2+y^2 = 39.0625$ by the distance from $0$ being $0.625$, and $3x+4y = 25$. Solving, you get two points : $(x,y) = (0,6.25)$, and $(x,y) = (6,1.75)$. These are the two candidate points for $D$.
POST-EDIT: If now you have understood, then I am willing to clarify two things. One, you might be wondering if the calculation to find $AD = 6.25$ is difficult. After all, we were (doing the equivalent of) squaring $3$ digit numbers there, and some six-digit arithmetic. But in fact, that part is really easy, and I can tell you why.
Secondly, how did I solve the equations $x^2+y^2 = 39.0625$ and $3x+4y=25$? That too was simple,and I can explain that as well if you have not understood yet.
POST-POST-EDIT : To solve this equation, we eliminate $y$ via $4y = 25-3x$. Multiplying the first equation by sixteen gives $16x^2+16y^2 = 625$, so $16x^2+(25-3x)^2 = 625$, a quadratic equation in $x$. From here it should be easy.
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
$begingroup$
I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
|
show 2 more comments
$begingroup$
The slope of $AC$ has been calculated wrong : it is in fact $frac{3-0}{4-0} = frac 34$. Hence, the slope of the line perpendicular to $AC$ must be $frac{-4}{3}$. So any point $D$ is located on the straight line passing through $C$ and having slope $frac{-4}{3}$ i.e. $(y-3) = -frac 43(x-4)$. This simplifies to $3y+4x = 25$.
Hence, $D$ satisfies these coordinates. Furthermore, the length of $AC$ is $5$, as you have said above.
Try to imagine the triangle ACD in your mind. This triangle is right angled at $C$, and the length of $AC$ is $5$.
EDIT : What is the definition of tangent of $DAC$? It is defined as the ratio between the opposite and adjacent sides, right? That is, $tan DAC = frac{OPP}{ADJ}$, where $OPP$ is the side opposite the angle $DAC$, and $ADJ$ is the side adjacent to $DAC$ which is not the hypotenuse (for that, we use the phrase $HYP$, so this side is not the hypotenuse).
Picturing the triangle in your mind, you must be able to see that $OPP = DC$, and $ADJ = AC$. Hence, $tan DAC = frac{DC}{AC} = frac 34$. Therefore, $DC = frac{3AC}{4} = frac{3 times 5}{4} = frac{15}{4} = 3.75$.
Now, you have a right angled triangle, hence by Pythagoras' theorem, $AD^2 = AC^2+DC^2$ (from your diagram, you must have seen that $AD$ is opposite angle $C$ which is right angled, so it is the hypotenuse). From this, $AD^2 = 25 + 14.0625 = 39.0625$, hence $AD = 6.25$.
So, if $D= (x,y)$, then $x^2+y^2 = 39.0625$ by the distance from $0$ being $0.625$, and $3x+4y = 25$. Solving, you get two points : $(x,y) = (0,6.25)$, and $(x,y) = (6,1.75)$. These are the two candidate points for $D$.
POST-EDIT: If now you have understood, then I am willing to clarify two things. One, you might be wondering if the calculation to find $AD = 6.25$ is difficult. After all, we were (doing the equivalent of) squaring $3$ digit numbers there, and some six-digit arithmetic. But in fact, that part is really easy, and I can tell you why.
Secondly, how did I solve the equations $x^2+y^2 = 39.0625$ and $3x+4y=25$? That too was simple,and I can explain that as well if you have not understood yet.
POST-POST-EDIT : To solve this equation, we eliminate $y$ via $4y = 25-3x$. Multiplying the first equation by sixteen gives $16x^2+16y^2 = 625$, so $16x^2+(25-3x)^2 = 625$, a quadratic equation in $x$. From here it should be easy.
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
$begingroup$
I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
|
show 2 more comments
$begingroup$
The slope of $AC$ has been calculated wrong : it is in fact $frac{3-0}{4-0} = frac 34$. Hence, the slope of the line perpendicular to $AC$ must be $frac{-4}{3}$. So any point $D$ is located on the straight line passing through $C$ and having slope $frac{-4}{3}$ i.e. $(y-3) = -frac 43(x-4)$. This simplifies to $3y+4x = 25$.
Hence, $D$ satisfies these coordinates. Furthermore, the length of $AC$ is $5$, as you have said above.
Try to imagine the triangle ACD in your mind. This triangle is right angled at $C$, and the length of $AC$ is $5$.
EDIT : What is the definition of tangent of $DAC$? It is defined as the ratio between the opposite and adjacent sides, right? That is, $tan DAC = frac{OPP}{ADJ}$, where $OPP$ is the side opposite the angle $DAC$, and $ADJ$ is the side adjacent to $DAC$ which is not the hypotenuse (for that, we use the phrase $HYP$, so this side is not the hypotenuse).
Picturing the triangle in your mind, you must be able to see that $OPP = DC$, and $ADJ = AC$. Hence, $tan DAC = frac{DC}{AC} = frac 34$. Therefore, $DC = frac{3AC}{4} = frac{3 times 5}{4} = frac{15}{4} = 3.75$.
Now, you have a right angled triangle, hence by Pythagoras' theorem, $AD^2 = AC^2+DC^2$ (from your diagram, you must have seen that $AD$ is opposite angle $C$ which is right angled, so it is the hypotenuse). From this, $AD^2 = 25 + 14.0625 = 39.0625$, hence $AD = 6.25$.
So, if $D= (x,y)$, then $x^2+y^2 = 39.0625$ by the distance from $0$ being $0.625$, and $3x+4y = 25$. Solving, you get two points : $(x,y) = (0,6.25)$, and $(x,y) = (6,1.75)$. These are the two candidate points for $D$.
POST-EDIT: If now you have understood, then I am willing to clarify two things. One, you might be wondering if the calculation to find $AD = 6.25$ is difficult. After all, we were (doing the equivalent of) squaring $3$ digit numbers there, and some six-digit arithmetic. But in fact, that part is really easy, and I can tell you why.
Secondly, how did I solve the equations $x^2+y^2 = 39.0625$ and $3x+4y=25$? That too was simple,and I can explain that as well if you have not understood yet.
POST-POST-EDIT : To solve this equation, we eliminate $y$ via $4y = 25-3x$. Multiplying the first equation by sixteen gives $16x^2+16y^2 = 625$, so $16x^2+(25-3x)^2 = 625$, a quadratic equation in $x$. From here it should be easy.
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
The slope of $AC$ has been calculated wrong : it is in fact $frac{3-0}{4-0} = frac 34$. Hence, the slope of the line perpendicular to $AC$ must be $frac{-4}{3}$. So any point $D$ is located on the straight line passing through $C$ and having slope $frac{-4}{3}$ i.e. $(y-3) = -frac 43(x-4)$. This simplifies to $3y+4x = 25$.
Hence, $D$ satisfies these coordinates. Furthermore, the length of $AC$ is $5$, as you have said above.
Try to imagine the triangle ACD in your mind. This triangle is right angled at $C$, and the length of $AC$ is $5$.
EDIT : What is the definition of tangent of $DAC$? It is defined as the ratio between the opposite and adjacent sides, right? That is, $tan DAC = frac{OPP}{ADJ}$, where $OPP$ is the side opposite the angle $DAC$, and $ADJ$ is the side adjacent to $DAC$ which is not the hypotenuse (for that, we use the phrase $HYP$, so this side is not the hypotenuse).
Picturing the triangle in your mind, you must be able to see that $OPP = DC$, and $ADJ = AC$. Hence, $tan DAC = frac{DC}{AC} = frac 34$. Therefore, $DC = frac{3AC}{4} = frac{3 times 5}{4} = frac{15}{4} = 3.75$.
Now, you have a right angled triangle, hence by Pythagoras' theorem, $AD^2 = AC^2+DC^2$ (from your diagram, you must have seen that $AD$ is opposite angle $C$ which is right angled, so it is the hypotenuse). From this, $AD^2 = 25 + 14.0625 = 39.0625$, hence $AD = 6.25$.
So, if $D= (x,y)$, then $x^2+y^2 = 39.0625$ by the distance from $0$ being $0.625$, and $3x+4y = 25$. Solving, you get two points : $(x,y) = (0,6.25)$, and $(x,y) = (6,1.75)$. These are the two candidate points for $D$.
POST-EDIT: If now you have understood, then I am willing to clarify two things. One, you might be wondering if the calculation to find $AD = 6.25$ is difficult. After all, we were (doing the equivalent of) squaring $3$ digit numbers there, and some six-digit arithmetic. But in fact, that part is really easy, and I can tell you why.
Secondly, how did I solve the equations $x^2+y^2 = 39.0625$ and $3x+4y=25$? That too was simple,and I can explain that as well if you have not understood yet.
POST-POST-EDIT : To solve this equation, we eliminate $y$ via $4y = 25-3x$. Multiplying the first equation by sixteen gives $16x^2+16y^2 = 625$, so $16x^2+(25-3x)^2 = 625$, a quadratic equation in $x$. From here it should be easy.
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
edited Jan 16 at 4:46
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
answered Sep 15 '17 at 1:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
38.6k33376
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
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I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
|
show 2 more comments
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
$begingroup$
I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
$begingroup$
I don't understand why DC = 3.75 and AD = 6.25. Can you please explain that more?
$endgroup$
– geo_freak
Sep 15 '17 at 1:57
$begingroup$
I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I will edit the answer to explain it.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:19
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
I have edited my answer. Please do have a look.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:29
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Thank you very much for the greatly-detailed explanation. I just have one question regarding how you found the measure of DC: why did you do $frac{3AC}{4}$ to find DC?
$endgroup$
– geo_freak
Sep 15 '17 at 2:39
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
$begingroup$
Since $frac{DC}{AC} = frac 34$, we cross multiply to get $4DC = 3AC$, and divide by $4$ on both sides to get $DC = frac{3AC}{4}$. Is this fine? If not, do get back.
$endgroup$
– астон вілла олоф мэллбэрг
Sep 15 '17 at 2:41
|
show 2 more comments
$begingroup$
Hint:)
Since slope of $AC$ is $dfrac34$ and $tan DAC=dfrac34$, use $tan(x+y)=dfrac{tan x+tan y}{1-tan xtan y}$ for finding the slope of $AD$. Find the intersection of lines $AD$ and $DC$.
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
Hint:)
Since slope of $AC$ is $dfrac34$ and $tan DAC=dfrac34$, use $tan(x+y)=dfrac{tan x+tan y}{1-tan xtan y}$ for finding the slope of $AD$. Find the intersection of lines $AD$ and $DC$.
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
Hint:)
Since slope of $AC$ is $dfrac34$ and $tan DAC=dfrac34$, use $tan(x+y)=dfrac{tan x+tan y}{1-tan xtan y}$ for finding the slope of $AD$. Find the intersection of lines $AD$ and $DC$.
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
$endgroup$
Hint:)
Since slope of $AC$ is $dfrac34$ and $tan DAC=dfrac34$, use $tan(x+y)=dfrac{tan x+tan y}{1-tan xtan y}$ for finding the slope of $AD$. Find the intersection of lines $AD$ and $DC$.
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
answered Sep 15 '17 at 1:50
NosratiNosrati
26.5k62354
26.5k62354
add a comment |
add a comment |
$begingroup$
You should sketch it:
$hspace{3cm}$
Using the distance formula, you can find:
$$AC=sqrt{(4-0)^2+(3-0)^2}=5 text{(which is basically Pythagoras formula)}.$$
Given $tan angle D_1AC=tan angle D_2AC=frac34$, you can find:
$$tan angle D_1AC=frac{CD_1}{AC}=frac34 Rightarrow CD_1=frac{3AC}{4}=frac{15}4=CD_2.$$
Note that the triangles $ABC$ and $BCE$ are similar, hence:
$$frac{CE}{AC}=frac{BC}{AB} Rightarrow frac{CE}{5}=frac{3}{4} Rightarrow CE=frac{15}{4}=CD_2 Rightarrow E=D_2;\
frac{BE}{BC}=frac{BC}{AB} Rightarrow BE=frac{BC^2}{AB}=frac{9}{4} Rightarrow D_2left(frac{25}{4},0right)$$
Now you can use the similarity of the triangles $BCE$ and $FD_1E$ to find the coordinates of $D_1$. It is an exercise for you. Answer is:
$D_1left(frac74,6right)$
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
You should sketch it:
$hspace{3cm}$
Using the distance formula, you can find:
$$AC=sqrt{(4-0)^2+(3-0)^2}=5 text{(which is basically Pythagoras formula)}.$$
Given $tan angle D_1AC=tan angle D_2AC=frac34$, you can find:
$$tan angle D_1AC=frac{CD_1}{AC}=frac34 Rightarrow CD_1=frac{3AC}{4}=frac{15}4=CD_2.$$
Note that the triangles $ABC$ and $BCE$ are similar, hence:
$$frac{CE}{AC}=frac{BC}{AB} Rightarrow frac{CE}{5}=frac{3}{4} Rightarrow CE=frac{15}{4}=CD_2 Rightarrow E=D_2;\
frac{BE}{BC}=frac{BC}{AB} Rightarrow BE=frac{BC^2}{AB}=frac{9}{4} Rightarrow D_2left(frac{25}{4},0right)$$
Now you can use the similarity of the triangles $BCE$ and $FD_1E$ to find the coordinates of $D_1$. It is an exercise for you. Answer is:
$D_1left(frac74,6right)$
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
You should sketch it:
$hspace{3cm}$
Using the distance formula, you can find:
$$AC=sqrt{(4-0)^2+(3-0)^2}=5 text{(which is basically Pythagoras formula)}.$$
Given $tan angle D_1AC=tan angle D_2AC=frac34$, you can find:
$$tan angle D_1AC=frac{CD_1}{AC}=frac34 Rightarrow CD_1=frac{3AC}{4}=frac{15}4=CD_2.$$
Note that the triangles $ABC$ and $BCE$ are similar, hence:
$$frac{CE}{AC}=frac{BC}{AB} Rightarrow frac{CE}{5}=frac{3}{4} Rightarrow CE=frac{15}{4}=CD_2 Rightarrow E=D_2;\
frac{BE}{BC}=frac{BC}{AB} Rightarrow BE=frac{BC^2}{AB}=frac{9}{4} Rightarrow D_2left(frac{25}{4},0right)$$
Now you can use the similarity of the triangles $BCE$ and $FD_1E$ to find the coordinates of $D_1$. It is an exercise for you. Answer is:
$D_1left(frac74,6right)$
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
$endgroup$
You should sketch it:
$hspace{3cm}$
Using the distance formula, you can find:
$$AC=sqrt{(4-0)^2+(3-0)^2}=5 text{(which is basically Pythagoras formula)}.$$
Given $tan angle D_1AC=tan angle D_2AC=frac34$, you can find:
$$tan angle D_1AC=frac{CD_1}{AC}=frac34 Rightarrow CD_1=frac{3AC}{4}=frac{15}4=CD_2.$$
Note that the triangles $ABC$ and $BCE$ are similar, hence:
$$frac{CE}{AC}=frac{BC}{AB} Rightarrow frac{CE}{5}=frac{3}{4} Rightarrow CE=frac{15}{4}=CD_2 Rightarrow E=D_2;\
frac{BE}{BC}=frac{BC}{AB} Rightarrow BE=frac{BC^2}{AB}=frac{9}{4} Rightarrow D_2left(frac{25}{4},0right)$$
Now you can use the similarity of the triangles $BCE$ and $FD_1E$ to find the coordinates of $D_1$. It is an exercise for you. Answer is:
$D_1left(frac74,6right)$
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
answered Jan 16 at 5:43
farruhotafarruhota
20.4k2739
20.4k2739
add a comment |
add a comment |
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$begingroup$
the slope of AC is not 5
$endgroup$
– Nosrati
Sep 15 '17 at 1:35
$begingroup$
@MyGlasses Ah yes, I found the distance of AC, not the slope! But I still do not know what to do with the tangent part of the problem.
$endgroup$
– geo_freak
Sep 15 '17 at 1:41