Mixing
let ABC be a right triangle at A such that $BC=2AB$. Find $angle ACB$
$begingroup$
So let $triangle ABC$ be a right triangle at vertex $A$ such that $BC=2AB$. Find the $angle ACB$
How can I find that angle without using cosine, sine and other things?
Since I've already figure out how to find it using cos: here's my approach:
We denote $angle ABC$ as $alpha$ so $$cosalpha=frac{AB}{BC}=frac{AB}{2AB}=frac{1}{2}$$
We do $cos^{-1}$ to find $angle ABC$ then we do $90^circ-angle ABC$ to find $angle ACB$.
So I'm looking for alternative way.
Thanks!
geometry
edited Jan 27 at 12:43
Larry
2,53531131
asked Mar 29 '14 at 13:09
user138849user138849
153
$endgroup$
add a comment |
$begingroup$
So let $triangle ABC$ be a right triangle at vertex $A$ such that $BC=2AB$. Find the $angle ACB$
How can I find that angle without using cosine, sine and other things?
Since I've already figure out how to find it using cos: here's my approach:
We denote $angle ABC$ as $alpha$ so $$cosalpha=frac{AB}{BC}=frac{AB}{2AB}=frac{1}{2}$$
We do $cos^{-1}$ to find $angle ABC$ then we do $90^circ-angle ABC$ to find $angle ACB$.
So I'm looking for alternative way.
Thanks!
geometry
edited Jan 27 at 12:43
Larry
2,53531131
asked Mar 29 '14 at 13:09
user138849user138849
153
$endgroup$
add a comment |
$begingroup$
So let $triangle ABC$ be a right triangle at vertex $A$ such that $BC=2AB$. Find the $angle ACB$
How can I find that angle without using cosine, sine and other things?
Since I've already figure out how to find it using cos: here's my approach:
We denote $angle ABC$ as $alpha$ so $$cosalpha=frac{AB}{BC}=frac{AB}{2AB}=frac{1}{2}$$
We do $cos^{-1}$ to find $angle ABC$ then we do $90^circ-angle ABC$ to find $angle ACB$.
So I'm looking for alternative way.
Thanks!
geometry
edited Jan 27 at 12:43
Larry
2,53531131
asked Mar 29 '14 at 13:09
user138849user138849
153
$endgroup$
So let $triangle ABC$ be a right triangle at vertex $A$ such that $BC=2AB$. Find the $angle ACB$
How can I find that angle without using cosine, sine and other things?
Since I've already figure out how to find it using cos: here's my approach:
We denote $angle ABC$ as $alpha$ so $$cosalpha=frac{AB}{BC}=frac{AB}{2AB}=frac{1}{2}$$
We do $cos^{-1}$ to find $angle ABC$ then we do $90^circ-angle ABC$ to find $angle ACB$.
So I'm looking for alternative way.
Thanks!
geometry
geometry
edited Jan 27 at 12:43
Larry
2,53531131
asked Mar 29 '14 at 13:09
user138849user138849
153
edited Jan 27 at 12:43
Larry
2,53531131
asked Mar 29 '14 at 13:09
user138849user138849
153
edited Jan 27 at 12:43
Larry
2,53531131
edited Jan 27 at 12:43
Larry
2,53531131
edited Jan 27 at 12:43
Larry
2,53531131
2,53531131
asked Mar 29 '14 at 13:09
user138849user138849
153
asked Mar 29 '14 at 13:09
user138849user138849
153
asked Mar 29 '14 at 13:09
user138849user138849
153
153
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
1
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
1
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
add a comment |
$begingroup$
Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
1
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
add a comment |
$begingroup$
Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
$endgroup$
Hint: take an equilateral triangle and draw one of the altitudes. Remember this altitude is also an angle bisector and a median, so...
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
answered Mar 29 '14 at 13:24
DonAntonioDonAntonio
180k1494233
180k1494233
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
1
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
add a comment |
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
1
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
I don't understand those terms, could you clarify a bit more please sir :)
$endgroup$
– user138849
Mar 29 '14 at 13:40
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
Altitude=height=segment of straight line from a vertex to the opposite side (a cevian) which is perpendicular to that side. What else isn't clear, @user138849 ?
$endgroup$
– DonAntonio
Mar 29 '14 at 13:42
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
I'm not a native english speaker so I don't know what equilateral triangle means, bisector, median
$endgroup$
– user138849
Mar 29 '14 at 13:52
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
$begingroup$
Then I can't help you, @user138849...I don't even know what your mother tongue is. If you ask in english then try to look for the corresponding terms in english.
$endgroup$
– DonAntonio
Mar 29 '14 at 13:56
1
1
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
$begingroup$
ookay i understand thanks anyway
$endgroup$
– user138849
Mar 29 '14 at 13:57
add a comment |
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