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How to construct a triangle with $BC=7.5$ cm. $angle ABC$=$60$° and $AC-AB=1.5$ cm.
How to construct a triangle with $BC=7.5$ cm.
$angle ABC$=$60$° and $AC-AB=1.5$ cm.
At first I constructed $BC$ then $angle ABC$ ,but I don't know what to do next. Please help me.
geometry geometric-construction
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
|
show 1 more comment
How to construct a triangle with $BC=7.5$ cm.
$angle ABC$=$60$° and $AC-AB=1.5$ cm.
At first I constructed $BC$ then $angle ABC$ ,but I don't know what to do next. Please help me.
geometry geometric-construction
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
2
How about utilising the cosine rule, where $AC=AB+1.5$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:03
@Raptor please explain more. Give me more hint. Cosine rule over which triangle?
– Sufaid Saleel
Nov 20 '18 at 15:05
2
The triangle $triangle ABC$, so $BC^2+AB^2-2 cdot AB cdot BCcos angle ABC=AC^2$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:08
Thanks! I have done the problem!
– Sufaid Saleel
Nov 20 '18 at 15:11
@Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant.
– Oldboy
Nov 20 '18 at 15:27
|
show 1 more comment
How to construct a triangle with $BC=7.5$ cm.
$angle ABC$=$60$° and $AC-AB=1.5$ cm.
At first I constructed $BC$ then $angle ABC$ ,but I don't know what to do next. Please help me.
geometry geometric-construction
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
How to construct a triangle with $BC=7.5$ cm.
$angle ABC$=$60$° and $AC-AB=1.5$ cm.
At first I constructed $BC$ then $angle ABC$ ,but I don't know what to do next. Please help me.
geometry geometric-construction
geometry geometric-construction
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
asked Nov 20 '18 at 15:00
Sufaid Saleel
1,757828
1,757828
2
How about utilising the cosine rule, where $AC=AB+1.5$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:03
@Raptor please explain more. Give me more hint. Cosine rule over which triangle?
– Sufaid Saleel
Nov 20 '18 at 15:05
2
The triangle $triangle ABC$, so $BC^2+AB^2-2 cdot AB cdot BCcos angle ABC=AC^2$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:08
Thanks! I have done the problem!
– Sufaid Saleel
Nov 20 '18 at 15:11
@Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant.
– Oldboy
Nov 20 '18 at 15:27
|
show 1 more comment
2
How about utilising the cosine rule, where $AC=AB+1.5$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:03
@Raptor please explain more. Give me more hint. Cosine rule over which triangle?
– Sufaid Saleel
Nov 20 '18 at 15:05
2
The triangle $triangle ABC$, so $BC^2+AB^2-2 cdot AB cdot BCcos angle ABC=AC^2$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:08
Thanks! I have done the problem!
– Sufaid Saleel
Nov 20 '18 at 15:11
@Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant.
– Oldboy
Nov 20 '18 at 15:27
2
2
How about utilising the cosine rule, where $AC=AB+1.5$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:03
How about utilising the cosine rule, where $AC=AB+1.5$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:03
@Raptor please explain more. Give me more hint. Cosine rule over which triangle?
– Sufaid Saleel
Nov 20 '18 at 15:05
@Raptor please explain more. Give me more hint. Cosine rule over which triangle?
– Sufaid Saleel
Nov 20 '18 at 15:05
2
2
The triangle $triangle ABC$, so $BC^2+AB^2-2 cdot AB cdot BCcos angle ABC=AC^2$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:08
The triangle $triangle ABC$, so $BC^2+AB^2-2 cdot AB cdot BCcos angle ABC=AC^2$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:08
Thanks! I have done the problem!
– Sufaid Saleel
Nov 20 '18 at 15:11
Thanks! I have done the problem!
– Sufaid Saleel
Nov 20 '18 at 15:11
@Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant.
– Oldboy
Nov 20 '18 at 15:27
@Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant.
– Oldboy
Nov 20 '18 at 15:27
|
show 1 more comment
1 Answer
1
active
oldest
votes
Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.
Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $angle CBC'=180^circ-angle ABC=120^circ$.
Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.
Problems like this one do not need trigonometry at all.
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.
Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $angle CBC'=180^circ-angle ABC=120^circ$.
Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.
Problems like this one do not need trigonometry at all.
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
add a comment |
Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.
Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $angle CBC'=180^circ-angle ABC=120^circ$.
Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.
Problems like this one do not need trigonometry at all.
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
add a comment |
Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.
Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $angle CBC'=180^circ-angle ABC=120^circ$.
Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.
Problems like this one do not need trigonometry at all.
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.
Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $angle CBC'=180^circ-angle ABC=120^circ$.
Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.
Problems like this one do not need trigonometry at all.
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
answered Nov 20 '18 at 15:26
Oldboy
6,9271832
6,9271832
add a comment |
add a comment |
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2
How about utilising the cosine rule, where $AC=AB+1.5$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:03
@Raptor please explain more. Give me more hint. Cosine rule over which triangle?
– Sufaid Saleel
Nov 20 '18 at 15:05
2
The triangle $triangle ABC$, so $BC^2+AB^2-2 cdot AB cdot BCcos angle ABC=AC^2$
– Mohammad Zuhair Khan
Nov 20 '18 at 15:08
Thanks! I have done the problem!
– Sufaid Saleel
Nov 20 '18 at 15:11
@Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant.
– Oldboy
Nov 20 '18 at 15:27