Mixing
Find out the angle of <ABC
$begingroup$
Help me to solve it please.how could it be done.I tried but nothing comes out.Help me please
geometry trigonometry triangle
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
$endgroup$
add a comment |
$begingroup$
Help me to solve it please.how could it be done.I tried but nothing comes out.Help me please
geometry trigonometry triangle
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
$endgroup$
$begingroup$
is trigonometry allowed?
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:49
$begingroup$
See $#132$ of qbyte.org/puzzles/puzzle14.html
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:51
$begingroup$
Drop a perpendicular from $A$ to $CB$ and notice characteristic 90-45-45 (half of a square)and 90-60-30 (half of equilateral triangle) triangles. Since the angle at $B$ is obviously $>90^circ$, try dividing it somehow to two angles ($30^circ+x$, $45^circ+x$, $90^circ+x$, etc).
$endgroup$
– Lazar Ljubenović
Nov 22 '13 at 8:11
add a comment |
$begingroup$
Help me to solve it please.how could it be done.I tried but nothing comes out.Help me please
geometry trigonometry triangle
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
$endgroup$
Help me to solve it please.how could it be done.I tried but nothing comes out.Help me please
geometry trigonometry triangle
geometry trigonometry triangle
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
edited Nov 22 '13 at 7:57
Fazla Rabbi Mashrur
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
asked Nov 22 '13 at 6:40
Fazla Rabbi MashrurFazla Rabbi Mashrur
918517
918517
$begingroup$
is trigonometry allowed?
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:49
$begingroup$
See $#132$ of qbyte.org/puzzles/puzzle14.html
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:51
$begingroup$
Drop a perpendicular from $A$ to $CB$ and notice characteristic 90-45-45 (half of a square)and 90-60-30 (half of equilateral triangle) triangles. Since the angle at $B$ is obviously $>90^circ$, try dividing it somehow to two angles ($30^circ+x$, $45^circ+x$, $90^circ+x$, etc).
$endgroup$
– Lazar Ljubenović
Nov 22 '13 at 8:11
add a comment |
$begingroup$
is trigonometry allowed?
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:49
$begingroup$
See $#132$ of qbyte.org/puzzles/puzzle14.html
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:51
$begingroup$
Drop a perpendicular from $A$ to $CB$ and notice characteristic 90-45-45 (half of a square)and 90-60-30 (half of equilateral triangle) triangles. Since the angle at $B$ is obviously $>90^circ$, try dividing it somehow to two angles ($30^circ+x$, $45^circ+x$, $90^circ+x$, etc).
$endgroup$
– Lazar Ljubenović
Nov 22 '13 at 8:11
$begingroup$
is trigonometry allowed?
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:49
$begingroup$
is trigonometry allowed?
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:49
$begingroup$
See $#132$ of qbyte.org/puzzles/puzzle14.html
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:51
$begingroup$
See $#132$ of qbyte.org/puzzles/puzzle14.html
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:51
$begingroup$
Drop a perpendicular from $A$ to $CB$ and notice characteristic 90-45-45 (half of a square)and 90-60-30 (half of equilateral triangle) triangles. Since the angle at $B$ is obviously $>90^circ$, try dividing it somehow to two angles ($30^circ+x$, $45^circ+x$, $90^circ+x$, etc).
$endgroup$
– Lazar Ljubenović
Nov 22 '13 at 8:11
$begingroup$
Drop a perpendicular from $A$ to $CB$ and notice characteristic 90-45-45 (half of a square)and 90-60-30 (half of equilateral triangle) triangles. Since the angle at $B$ is obviously $>90^circ$, try dividing it somehow to two angles ($30^circ+x$, $45^circ+x$, $90^circ+x$, etc).
$endgroup$
– Lazar Ljubenović
Nov 22 '13 at 8:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Let $angle BAD = alpha$, so by sine rule,
$$frac{sin alpha}{BD}=frac{sin 45^{circ}}{AB}$$
and
$$frac{sin (alpha+ 15^{circ})}{2BD}=frac{sin 30^{circ}}{AB}$$
so
$$frac{2sin alpha}{sin (alpha+ 15^{circ})}=frac{sin 45^{circ}}{sin 30^{circ}}$$
After finding $alpha$, you can then find $angle ABC$
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
$endgroup$
add a comment |
$begingroup$
we draw BO$perp$AC and we join OD
- see in $bigtriangleup$BOC $angle$BOC=90 and OD is midian. so OD=BD=DC
- NOW $angle$OCD=30 so $angle$OBD=60=$angle$NOD[ where BO meets AD at N]
- now $angle$NBD=60 and $angle$NDB=45 so $angle$BND=75
- $angle$BND=75 and $angle$NOD=60 so $angle$ODN=15=$angle$OAD. so AO=OD
- now $angle$ODB=45+15=60. so $bigtriangleup$BOD is equilateral. so OD=BO =BD AND AO=OD
- so we have AO=BO and we see $angle$AOB=90 so $angle$OAB=45
- we name $angle$DAB=x so x+15 =45 so x=30
- in $bigtriangleup$ABD $angle$DAB=30 and $angle$ADB=45 so $angle$ABC=105
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
$endgroup$
add a comment |
$begingroup$
Do construction as shown 
WLOG, we can let BD = DC = 1.
Applying special angle relations to $triangle AHC$, we have
$h : h + 1 = 1 : sqrt{3} leftrightarrow h = frac{sqrt(3) +1} {2}$
Then calculate $h - 1$ for later use. $h – 1 = frac {sqrt(3) – 1} {2}$
In $triangle AHB, AB^2 = h^2 + (h – 1)^2 leftrightarrow AB = sqrt 2$
Then, $dfrac {AB}{BD} = dfrac {sqrt 2}{1}$ and $dfrac {BC}{AB} = dfrac {2}{sqrt 2} = dfrac {sqrt 2}{1}$.
By ratio of 2 sides and the included angle, $triangle ABCsimtriangle DBA$.
Therefore, $angle BAD = angle BCA = 30^{circ}$
Hence $angle ABC = 105^{circ}$
Note:- This figure is commonly used in calculating sin (15).
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $angle BAD = alpha$, so by sine rule,
$$frac{sin alpha}{BD}=frac{sin 45^{circ}}{AB}$$
and
$$frac{sin (alpha+ 15^{circ})}{2BD}=frac{sin 30^{circ}}{AB}$$
so
$$frac{2sin alpha}{sin (alpha+ 15^{circ})}=frac{sin 45^{circ}}{sin 30^{circ}}$$
After finding $alpha$, you can then find $angle ABC$
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
$endgroup$
add a comment |
$begingroup$
Hint: Let $angle BAD = alpha$, so by sine rule,
$$frac{sin alpha}{BD}=frac{sin 45^{circ}}{AB}$$
and
$$frac{sin (alpha+ 15^{circ})}{2BD}=frac{sin 30^{circ}}{AB}$$
so
$$frac{2sin alpha}{sin (alpha+ 15^{circ})}=frac{sin 45^{circ}}{sin 30^{circ}}$$
After finding $alpha$, you can then find $angle ABC$
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
$endgroup$
add a comment |
$begingroup$
Hint: Let $angle BAD = alpha$, so by sine rule,
$$frac{sin alpha}{BD}=frac{sin 45^{circ}}{AB}$$
and
$$frac{sin (alpha+ 15^{circ})}{2BD}=frac{sin 30^{circ}}{AB}$$
so
$$frac{2sin alpha}{sin (alpha+ 15^{circ})}=frac{sin 45^{circ}}{sin 30^{circ}}$$
After finding $alpha$, you can then find $angle ABC$
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
$endgroup$
Hint: Let $angle BAD = alpha$, so by sine rule,
$$frac{sin alpha}{BD}=frac{sin 45^{circ}}{AB}$$
and
$$frac{sin (alpha+ 15^{circ})}{2BD}=frac{sin 30^{circ}}{AB}$$
so
$$frac{2sin alpha}{sin (alpha+ 15^{circ})}=frac{sin 45^{circ}}{sin 30^{circ}}$$
After finding $alpha$, you can then find $angle ABC$
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
answered Nov 22 '13 at 6:48
freak_warriorfreak_warrior
2,8651131
2,8651131
add a comment |
add a comment |
$begingroup$
we draw BO$perp$AC and we join OD
- see in $bigtriangleup$BOC $angle$BOC=90 and OD is midian. so OD=BD=DC
- NOW $angle$OCD=30 so $angle$OBD=60=$angle$NOD[ where BO meets AD at N]
- now $angle$NBD=60 and $angle$NDB=45 so $angle$BND=75
- $angle$BND=75 and $angle$NOD=60 so $angle$ODN=15=$angle$OAD. so AO=OD
- now $angle$ODB=45+15=60. so $bigtriangleup$BOD is equilateral. so OD=BO =BD AND AO=OD
- so we have AO=BO and we see $angle$AOB=90 so $angle$OAB=45
- we name $angle$DAB=x so x+15 =45 so x=30
- in $bigtriangleup$ABD $angle$DAB=30 and $angle$ADB=45 so $angle$ABC=105
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
$endgroup$
add a comment |
$begingroup$
we draw BO$perp$AC and we join OD
- see in $bigtriangleup$BOC $angle$BOC=90 and OD is midian. so OD=BD=DC
- NOW $angle$OCD=30 so $angle$OBD=60=$angle$NOD[ where BO meets AD at N]
- now $angle$NBD=60 and $angle$NDB=45 so $angle$BND=75
- $angle$BND=75 and $angle$NOD=60 so $angle$ODN=15=$angle$OAD. so AO=OD
- now $angle$ODB=45+15=60. so $bigtriangleup$BOD is equilateral. so OD=BO =BD AND AO=OD
- so we have AO=BO and we see $angle$AOB=90 so $angle$OAB=45
- we name $angle$DAB=x so x+15 =45 so x=30
- in $bigtriangleup$ABD $angle$DAB=30 and $angle$ADB=45 so $angle$ABC=105
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
$endgroup$
add a comment |
$begingroup$
we draw BO$perp$AC and we join OD
- see in $bigtriangleup$BOC $angle$BOC=90 and OD is midian. so OD=BD=DC
- NOW $angle$OCD=30 so $angle$OBD=60=$angle$NOD[ where BO meets AD at N]
- now $angle$NBD=60 and $angle$NDB=45 so $angle$BND=75
- $angle$BND=75 and $angle$NOD=60 so $angle$ODN=15=$angle$OAD. so AO=OD
- now $angle$ODB=45+15=60. so $bigtriangleup$BOD is equilateral. so OD=BO =BD AND AO=OD
- so we have AO=BO and we see $angle$AOB=90 so $angle$OAB=45
- we name $angle$DAB=x so x+15 =45 so x=30
- in $bigtriangleup$ABD $angle$DAB=30 and $angle$ADB=45 so $angle$ABC=105
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
$endgroup$
we draw BO$perp$AC and we join OD
- see in $bigtriangleup$BOC $angle$BOC=90 and OD is midian. so OD=BD=DC
- NOW $angle$OCD=30 so $angle$OBD=60=$angle$NOD[ where BO meets AD at N]
- now $angle$NBD=60 and $angle$NDB=45 so $angle$BND=75
- $angle$BND=75 and $angle$NOD=60 so $angle$ODN=15=$angle$OAD. so AO=OD
- now $angle$ODB=45+15=60. so $bigtriangleup$BOD is equilateral. so OD=BO =BD AND AO=OD
- so we have AO=BO and we see $angle$AOB=90 so $angle$OAB=45
- we name $angle$DAB=x so x+15 =45 so x=30
- in $bigtriangleup$ABD $angle$DAB=30 and $angle$ADB=45 so $angle$ABC=105
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
answered Nov 22 '13 at 8:03
krishan actonkrishan acton
420214
420214
add a comment |
add a comment |
$begingroup$
Do construction as shown
WLOG, we can let BD = DC = 1.
Applying special angle relations to $triangle AHC$, we have
$h : h + 1 = 1 : sqrt{3} leftrightarrow h = frac{sqrt(3) +1} {2}$
Then calculate $h - 1$ for later use. $h – 1 = frac {sqrt(3) – 1} {2}$
In $triangle AHB, AB^2 = h^2 + (h – 1)^2 leftrightarrow AB = sqrt 2$
Then, $dfrac {AB}{BD} = dfrac {sqrt 2}{1}$ and $dfrac {BC}{AB} = dfrac {2}{sqrt 2} = dfrac {sqrt 2}{1}$.
By ratio of 2 sides and the included angle, $triangle ABCsimtriangle DBA$.
Therefore, $angle BAD = angle BCA = 30^{circ}$
Hence $angle ABC = 105^{circ}$
Note:- This figure is commonly used in calculating sin (15).
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
Do construction as shown
WLOG, we can let BD = DC = 1.
Applying special angle relations to $triangle AHC$, we have
$h : h + 1 = 1 : sqrt{3} leftrightarrow h = frac{sqrt(3) +1} {2}$
Then calculate $h - 1$ for later use. $h – 1 = frac {sqrt(3) – 1} {2}$
In $triangle AHB, AB^2 = h^2 + (h – 1)^2 leftrightarrow AB = sqrt 2$
Then, $dfrac {AB}{BD} = dfrac {sqrt 2}{1}$ and $dfrac {BC}{AB} = dfrac {2}{sqrt 2} = dfrac {sqrt 2}{1}$.
By ratio of 2 sides and the included angle, $triangle ABCsimtriangle DBA$.
Therefore, $angle BAD = angle BCA = 30^{circ}$
Hence $angle ABC = 105^{circ}$
Note:- This figure is commonly used in calculating sin (15).
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
Do construction as shown
WLOG, we can let BD = DC = 1.
Applying special angle relations to $triangle AHC$, we have
$h : h + 1 = 1 : sqrt{3} leftrightarrow h = frac{sqrt(3) +1} {2}$
Then calculate $h - 1$ for later use. $h – 1 = frac {sqrt(3) – 1} {2}$
In $triangle AHB, AB^2 = h^2 + (h – 1)^2 leftrightarrow AB = sqrt 2$
Then, $dfrac {AB}{BD} = dfrac {sqrt 2}{1}$ and $dfrac {BC}{AB} = dfrac {2}{sqrt 2} = dfrac {sqrt 2}{1}$.
By ratio of 2 sides and the included angle, $triangle ABCsimtriangle DBA$.
Therefore, $angle BAD = angle BCA = 30^{circ}$
Hence $angle ABC = 105^{circ}$
Note:- This figure is commonly used in calculating sin (15).
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
$endgroup$
Do construction as shown
WLOG, we can let BD = DC = 1.
Applying special angle relations to $triangle AHC$, we have
$h : h + 1 = 1 : sqrt{3} leftrightarrow h = frac{sqrt(3) +1} {2}$
Then calculate $h - 1$ for later use. $h – 1 = frac {sqrt(3) – 1} {2}$
In $triangle AHB, AB^2 = h^2 + (h – 1)^2 leftrightarrow AB = sqrt 2$
Then, $dfrac {AB}{BD} = dfrac {sqrt 2}{1}$ and $dfrac {BC}{AB} = dfrac {2}{sqrt 2} = dfrac {sqrt 2}{1}$.
By ratio of 2 sides and the included angle, $triangle ABCsimtriangle DBA$.
Therefore, $angle BAD = angle BCA = 30^{circ}$
Hence $angle ABC = 105^{circ}$
Note:- This figure is commonly used in calculating sin (15).
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
edited Jan 20 at 13:52
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
answered Nov 22 '13 at 16:29
MickMick
11.9k21641
11.9k21641
add a comment |
add a comment |
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$begingroup$
is trigonometry allowed?
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:49
$begingroup$
See $#132$ of qbyte.org/puzzles/puzzle14.html
$endgroup$
– lab bhattacharjee
Nov 22 '13 at 6:51
$begingroup$
Drop a perpendicular from $A$ to $CB$ and notice characteristic 90-45-45 (half of a square)and 90-60-30 (half of equilateral triangle) triangles. Since the angle at $B$ is obviously $>90^circ$, try dividing it somehow to two angles ($30^circ+x$, $45^circ+x$, $90^circ+x$, etc).
$endgroup$
– Lazar Ljubenović
Nov 22 '13 at 8:11