Show that base is twice the height if base angles of a triangle are $22.5^circ$ and $112.5^circ$
$begingroup$
The base angles of a triangle are $22.5^circ$ and $112.5^circ$. Show that the base is twice the height.
My Attempt
$$
h=c.sin22.5^circ=c.cos 67.5^circ\
=bsin 67.5^circ=bcos 22.5^circ
$$
$$
a=ccos22.5^circ- bsin22.5circ=frac{h}{b}-frac{h}{c}=hcdotfrac{c-b}{bc}
$$
I have no clue of how to prove this.
trigonometry triangle
edited Jan 1 at 23:50
A.Γ.
22.7k32656
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
$endgroup$
add a comment |
$begingroup$
The base angles of a triangle are $22.5^circ$ and $112.5^circ$. Show that the base is twice the height.
My Attempt
$$
h=c.sin22.5^circ=c.cos 67.5^circ\
=bsin 67.5^circ=bcos 22.5^circ
$$
$$
a=ccos22.5^circ- bsin22.5circ=frac{h}{b}-frac{h}{c}=hcdotfrac{c-b}{bc}
$$
I have no clue of how to prove this.
trigonometry triangle
edited Jan 1 at 23:50
A.Γ.
22.7k32656
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
$endgroup$
$begingroup$
You should have pointed upon every vertex.
$endgroup$
– Rakibul Islam Prince
Jan 1 at 23:15
1
$begingroup$
@RakibulIslamPrince Fixed.
$endgroup$
– A.Γ.
Jan 1 at 23:51
add a comment |
$begingroup$
The base angles of a triangle are $22.5^circ$ and $112.5^circ$. Show that the base is twice the height.
My Attempt
$$
h=c.sin22.5^circ=c.cos 67.5^circ\
=bsin 67.5^circ=bcos 22.5^circ
$$
$$
a=ccos22.5^circ- bsin22.5circ=frac{h}{b}-frac{h}{c}=hcdotfrac{c-b}{bc}
$$
I have no clue of how to prove this.
trigonometry triangle
edited Jan 1 at 23:50
A.Γ.
22.7k32656
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
$endgroup$
The base angles of a triangle are $22.5^circ$ and $112.5^circ$. Show that the base is twice the height.
My Attempt
$$
h=c.sin22.5^circ=c.cos 67.5^circ\
=bsin 67.5^circ=bcos 22.5^circ
$$
$$
a=ccos22.5^circ- bsin22.5circ=frac{h}{b}-frac{h}{c}=hcdotfrac{c-b}{bc}
$$
I have no clue of how to prove this.
trigonometry triangle
trigonometry triangle
edited Jan 1 at 23:50
A.Γ.
22.7k32656
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
edited Jan 1 at 23:50
A.Γ.
22.7k32656
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
edited Jan 1 at 23:50
A.Γ.
22.7k32656
edited Jan 1 at 23:50
A.Γ.
22.7k32656
edited Jan 1 at 23:50
A.Γ.
22.7k32656
22.7k32656
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
asked Jan 1 at 22:48
ss1729ss1729
1,8991723
1,8991723
$begingroup$
You should have pointed upon every vertex.
$endgroup$
– Rakibul Islam Prince
Jan 1 at 23:15
1
$begingroup$
@RakibulIslamPrince Fixed.
$endgroup$
– A.Γ.
Jan 1 at 23:51
add a comment |
$begingroup$
You should have pointed upon every vertex.
$endgroup$
– Rakibul Islam Prince
Jan 1 at 23:15
1
$begingroup$
@RakibulIslamPrince Fixed.
$endgroup$
– A.Γ.
Jan 1 at 23:51
$begingroup$
You should have pointed upon every vertex.
$endgroup$
– Rakibul Islam Prince
Jan 1 at 23:15
$begingroup$
You should have pointed upon every vertex.
$endgroup$
– Rakibul Islam Prince
Jan 1 at 23:15
1
1
$begingroup$
@RakibulIslamPrince Fixed.
$endgroup$
– A.Γ.
Jan 1 at 23:51
$begingroup$
@RakibulIslamPrince Fixed.
$endgroup$
– A.Γ.
Jan 1 at 23:51
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
By the law of sines, $$frac{a}{sin{45^{circ}}}=frac{b}{sin{22.5^{circ}}}$$
By the double angle formula, this is equivalent to $$frac{a}{2sin{22.5^{circ}}cos{22.5^{circ}}}=frac{b}{sin{22.5^{circ}}}impliesfrac{a}{2cos{22.5^{circ}}}=b$$
From the smaller right triangle we see that $$frac{h}{b}=cos{22.5^{circ}}implies h=bcos{22.5^{circ}}$$
Combining the results gives $a=2h$.
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
$endgroup$
add a comment |
$begingroup$
The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$frac{h}{a+x}=frac xh=tan22.5=sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(sqrt{2}-1)+h^2(sqrt{2}-1)^2$$
Rearranging gives $$frac ah=frac{2sqrt{2}-2}{sqrt{2}-1}=2$$
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
$endgroup$
add a comment |
$begingroup$
Let,the extended portion of $a=x$ and let $D$ be the intersection point of base and height.
So,$CD=x$
Now,in the $triangle ABD$,
$$tan (22.5)=frac{h}{a+x}implies h=atan(22.5)+xtan(22.5)......(1)$$
in the $triangle ACD$,
$$tan (67.5)=frac{h}{x}implies h=xtan (67.5)implies x=frac{h}{tan (67.5)}$$
from (1),
$$h=atan(22.5)+frac{h}{tan (67.5)}tan(22.5)$$
$$implies h=frac{a}{2}[text{after simplification}]$$
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
$endgroup$
add a comment |
$begingroup$
The base of the triangle is $ccos(22.5)-bcos(180-112.5)$. The height is $csin(22.5)$. Also, $bsin(180-112.5) = csin(22.5)$. Use these two relations to write $b$ in terms of $c$ and show that $$frac{csin(22.5)}{ccos(22.5)-bcos(180-112.5)} = 1/2.$$ (Hint: the $c$ will cancel out).
answered Jan 1 at 22:56
D.B.D.B.
1,2128
$endgroup$
add a comment |
$begingroup$
Solution without trigonometry.
Since $$measuredangle CAD=measuredangle BCD-90^{circ}=112.5^{circ}-90^{circ}=22.5^{circ}=measuredangle ABD,$$
we get that $DA$ is a tangent line to the circumcircle of $Delta ABC$.
Let $O$ be a center of the circle and $OM$ be an altitude of $Delta OBC$.
Since $OAperp DA$, we obtain $$measuredangle $OCM=measuredangle ABC-measuredangle OCA=measuredangle ABC-measuredangle OAC=112.5^{circ}-(90^{circ}-22.5^{circ})=45^{circ},$$
which says $OM=MC$ and since $BM=MC,$ we obtain:
$$BC=2OM=2AD$$ and we are done!
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
add a comment |
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5 Answers
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$begingroup$
By the law of sines, $$frac{a}{sin{45^{circ}}}=frac{b}{sin{22.5^{circ}}}$$
By the double angle formula, this is equivalent to $$frac{a}{2sin{22.5^{circ}}cos{22.5^{circ}}}=frac{b}{sin{22.5^{circ}}}impliesfrac{a}{2cos{22.5^{circ}}}=b$$
From the smaller right triangle we see that $$frac{h}{b}=cos{22.5^{circ}}implies h=bcos{22.5^{circ}}$$
Combining the results gives $a=2h$.
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
$endgroup$
add a comment |
$begingroup$
By the law of sines, $$frac{a}{sin{45^{circ}}}=frac{b}{sin{22.5^{circ}}}$$
By the double angle formula, this is equivalent to $$frac{a}{2sin{22.5^{circ}}cos{22.5^{circ}}}=frac{b}{sin{22.5^{circ}}}impliesfrac{a}{2cos{22.5^{circ}}}=b$$
From the smaller right triangle we see that $$frac{h}{b}=cos{22.5^{circ}}implies h=bcos{22.5^{circ}}$$
Combining the results gives $a=2h$.
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
$endgroup$
add a comment |
$begingroup$
By the law of sines, $$frac{a}{sin{45^{circ}}}=frac{b}{sin{22.5^{circ}}}$$
By the double angle formula, this is equivalent to $$frac{a}{2sin{22.5^{circ}}cos{22.5^{circ}}}=frac{b}{sin{22.5^{circ}}}impliesfrac{a}{2cos{22.5^{circ}}}=b$$
From the smaller right triangle we see that $$frac{h}{b}=cos{22.5^{circ}}implies h=bcos{22.5^{circ}}$$
Combining the results gives $a=2h$.
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
$endgroup$
By the law of sines, $$frac{a}{sin{45^{circ}}}=frac{b}{sin{22.5^{circ}}}$$
By the double angle formula, this is equivalent to $$frac{a}{2sin{22.5^{circ}}cos{22.5^{circ}}}=frac{b}{sin{22.5^{circ}}}impliesfrac{a}{2cos{22.5^{circ}}}=b$$
From the smaller right triangle we see that $$frac{h}{b}=cos{22.5^{circ}}implies h=bcos{22.5^{circ}}$$
Combining the results gives $a=2h$.
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
answered Jan 1 at 23:31
John DoumaJohn Douma
5,53211319
5,53211319
add a comment |
add a comment |
$begingroup$
The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$frac{h}{a+x}=frac xh=tan22.5=sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(sqrt{2}-1)+h^2(sqrt{2}-1)^2$$
Rearranging gives $$frac ah=frac{2sqrt{2}-2}{sqrt{2}-1}=2$$
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
$endgroup$
add a comment |
$begingroup$
The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$frac{h}{a+x}=frac xh=tan22.5=sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(sqrt{2}-1)+h^2(sqrt{2}-1)^2$$
Rearranging gives $$frac ah=frac{2sqrt{2}-2}{sqrt{2}-1}=2$$
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
$endgroup$
add a comment |
$begingroup$
The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$frac{h}{a+x}=frac xh=tan22.5=sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(sqrt{2}-1)+h^2(sqrt{2}-1)^2$$
Rearranging gives $$frac ah=frac{2sqrt{2}-2}{sqrt{2}-1}=2$$
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
$endgroup$
The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$frac{h}{a+x}=frac xh=tan22.5=sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(sqrt{2}-1)+h^2(sqrt{2}-1)^2$$
Rearranging gives $$frac ah=frac{2sqrt{2}-2}{sqrt{2}-1}=2$$
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
answered Jan 1 at 23:47
David QuinnDavid Quinn
23.9k21141
23.9k21141
add a comment |
add a comment |
$begingroup$
Let,the extended portion of $a=x$ and let $D$ be the intersection point of base and height.
So,$CD=x$
Now,in the $triangle ABD$,
$$tan (22.5)=frac{h}{a+x}implies h=atan(22.5)+xtan(22.5)......(1)$$
in the $triangle ACD$,
$$tan (67.5)=frac{h}{x}implies h=xtan (67.5)implies x=frac{h}{tan (67.5)}$$
from (1),
$$h=atan(22.5)+frac{h}{tan (67.5)}tan(22.5)$$
$$implies h=frac{a}{2}[text{after simplification}]$$
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
$endgroup$
add a comment |
$begingroup$
Let,the extended portion of $a=x$ and let $D$ be the intersection point of base and height.
So,$CD=x$
Now,in the $triangle ABD$,
$$tan (22.5)=frac{h}{a+x}implies h=atan(22.5)+xtan(22.5)......(1)$$
in the $triangle ACD$,
$$tan (67.5)=frac{h}{x}implies h=xtan (67.5)implies x=frac{h}{tan (67.5)}$$
from (1),
$$h=atan(22.5)+frac{h}{tan (67.5)}tan(22.5)$$
$$implies h=frac{a}{2}[text{after simplification}]$$
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
$endgroup$
add a comment |
$begingroup$
Let,the extended portion of $a=x$ and let $D$ be the intersection point of base and height.
So,$CD=x$
Now,in the $triangle ABD$,
$$tan (22.5)=frac{h}{a+x}implies h=atan(22.5)+xtan(22.5)......(1)$$
in the $triangle ACD$,
$$tan (67.5)=frac{h}{x}implies h=xtan (67.5)implies x=frac{h}{tan (67.5)}$$
from (1),
$$h=atan(22.5)+frac{h}{tan (67.5)}tan(22.5)$$
$$implies h=frac{a}{2}[text{after simplification}]$$
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
$endgroup$
Let,the extended portion of $a=x$ and let $D$ be the intersection point of base and height.
So,$CD=x$
Now,in the $triangle ABD$,
$$tan (22.5)=frac{h}{a+x}implies h=atan(22.5)+xtan(22.5)......(1)$$
in the $triangle ACD$,
$$tan (67.5)=frac{h}{x}implies h=xtan (67.5)implies x=frac{h}{tan (67.5)}$$
from (1),
$$h=atan(22.5)+frac{h}{tan (67.5)}tan(22.5)$$
$$implies h=frac{a}{2}[text{after simplification}]$$
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
edited Jan 1 at 23:12
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
answered Jan 1 at 23:06
Rakibul Islam PrinceRakibul Islam Prince
1,008211
1,008211
add a comment |
add a comment |
$begingroup$
The base of the triangle is $ccos(22.5)-bcos(180-112.5)$. The height is $csin(22.5)$. Also, $bsin(180-112.5) = csin(22.5)$. Use these two relations to write $b$ in terms of $c$ and show that $$frac{csin(22.5)}{ccos(22.5)-bcos(180-112.5)} = 1/2.$$ (Hint: the $c$ will cancel out).
answered Jan 1 at 22:56
D.B.D.B.
1,2128
$endgroup$
add a comment |
$begingroup$
The base of the triangle is $ccos(22.5)-bcos(180-112.5)$. The height is $csin(22.5)$. Also, $bsin(180-112.5) = csin(22.5)$. Use these two relations to write $b$ in terms of $c$ and show that $$frac{csin(22.5)}{ccos(22.5)-bcos(180-112.5)} = 1/2.$$ (Hint: the $c$ will cancel out).
answered Jan 1 at 22:56
D.B.D.B.
1,2128
$endgroup$
add a comment |
$begingroup$
The base of the triangle is $ccos(22.5)-bcos(180-112.5)$. The height is $csin(22.5)$. Also, $bsin(180-112.5) = csin(22.5)$. Use these two relations to write $b$ in terms of $c$ and show that $$frac{csin(22.5)}{ccos(22.5)-bcos(180-112.5)} = 1/2.$$ (Hint: the $c$ will cancel out).
answered Jan 1 at 22:56
D.B.D.B.
1,2128
$endgroup$
The base of the triangle is $ccos(22.5)-bcos(180-112.5)$. The height is $csin(22.5)$. Also, $bsin(180-112.5) = csin(22.5)$. Use these two relations to write $b$ in terms of $c$ and show that $$frac{csin(22.5)}{ccos(22.5)-bcos(180-112.5)} = 1/2.$$ (Hint: the $c$ will cancel out).
answered Jan 1 at 22:56
D.B.D.B.
1,2128
answered Jan 1 at 22:56
D.B.D.B.
1,2128
answered Jan 1 at 22:56
D.B.D.B.
1,2128
answered Jan 1 at 22:56
D.B.D.B.
1,2128
1,2128
add a comment |
add a comment |
$begingroup$
Solution without trigonometry.
Since $$measuredangle CAD=measuredangle BCD-90^{circ}=112.5^{circ}-90^{circ}=22.5^{circ}=measuredangle ABD,$$
we get that $DA$ is a tangent line to the circumcircle of $Delta ABC$.
Let $O$ be a center of the circle and $OM$ be an altitude of $Delta OBC$.
Since $OAperp DA$, we obtain $$measuredangle $OCM=measuredangle ABC-measuredangle OCA=measuredangle ABC-measuredangle OAC=112.5^{circ}-(90^{circ}-22.5^{circ})=45^{circ},$$
which says $OM=MC$ and since $BM=MC,$ we obtain:
$$BC=2OM=2AD$$ and we are done!
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
add a comment |
$begingroup$
Solution without trigonometry.
Since $$measuredangle CAD=measuredangle BCD-90^{circ}=112.5^{circ}-90^{circ}=22.5^{circ}=measuredangle ABD,$$
we get that $DA$ is a tangent line to the circumcircle of $Delta ABC$.
Let $O$ be a center of the circle and $OM$ be an altitude of $Delta OBC$.
Since $OAperp DA$, we obtain $$measuredangle $OCM=measuredangle ABC-measuredangle OCA=measuredangle ABC-measuredangle OAC=112.5^{circ}-(90^{circ}-22.5^{circ})=45^{circ},$$
which says $OM=MC$ and since $BM=MC,$ we obtain:
$$BC=2OM=2AD$$ and we are done!
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
add a comment |
$begingroup$
Solution without trigonometry.
Since $$measuredangle CAD=measuredangle BCD-90^{circ}=112.5^{circ}-90^{circ}=22.5^{circ}=measuredangle ABD,$$
we get that $DA$ is a tangent line to the circumcircle of $Delta ABC$.
Let $O$ be a center of the circle and $OM$ be an altitude of $Delta OBC$.
Since $OAperp DA$, we obtain $$measuredangle $OCM=measuredangle ABC-measuredangle OCA=measuredangle ABC-measuredangle OAC=112.5^{circ}-(90^{circ}-22.5^{circ})=45^{circ},$$
which says $OM=MC$ and since $BM=MC,$ we obtain:
$$BC=2OM=2AD$$ and we are done!
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
Solution without trigonometry.
Since $$measuredangle CAD=measuredangle BCD-90^{circ}=112.5^{circ}-90^{circ}=22.5^{circ}=measuredangle ABD,$$
we get that $DA$ is a tangent line to the circumcircle of $Delta ABC$.
Let $O$ be a center of the circle and $OM$ be an altitude of $Delta OBC$.
Since $OAperp DA$, we obtain $$measuredangle $OCM=measuredangle ABC-measuredangle OCA=measuredangle ABC-measuredangle OAC=112.5^{circ}-(90^{circ}-22.5^{circ})=45^{circ},$$
which says $OM=MC$ and since $BM=MC,$ we obtain:
$$BC=2OM=2AD$$ and we are done!
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
answered Jan 2 at 4:31
Michael RozenbergMichael Rozenberg
99k1590189
99k1590189
add a comment |
add a comment |
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$begingroup$
You should have pointed upon every vertex.
$endgroup$
– Rakibul Islam Prince
Jan 1 at 23:15
1
$begingroup$
@RakibulIslamPrince Fixed.
$endgroup$
– A.Γ.
Jan 1 at 23:51