難解とか 著名らしい

質問

《あなをかし　法[と命名します]》; 下　紫枠の　穴　に　適するものを　挿入し

http://userdisk.webry.biglobe.ne.jp/020/691/47/N000/000/006/150923830756382814177.gif

角　t を　求めて下さい；

回答No.1

jcpmutura

直線BA
y=xtan(80°)
と直線AC
y=(x-5)tan(130°)
の交点A=(x,y)を求めると

xtan(80°)=(x-5)tan(130°)
x{tan(80°)-tan(130°)}=-5tan(130°)
x{sin(80°)/cos(80°)-sin(130°)/cos(130°)}=-5sin(130°)/cos(130°)
x{sin(80°)cos(130°)-sin(130°)cos(80°)}=-5sin(130°)cos(80°)
x{-sin(80°)cos(50°)-sin(50°)cos(80°)}=-5sin(50°)cos(80°)
x{sin(80°)cos(50°)+sin(50°)cos(80°)}=5sin(50°)cos(80°)
xsin(130°)=5sin(50°)cos(80°)
xsin(50°)=5sin(50°)cos(80°)
x=5cos(80°)
y=5tan(80°)cos(80°)
y=5sin(80°)
∴
A=(5cos(80°),5sin(80°))

直線BD
y=x√3
と直線DC
y=(x-5)tan(100°)
の交点D=(x,y)を求めると

x√3=(x-5)tan(100°)
x{√3-tan(100°)}=-5tan(100°)
x{√3-sin(100°)/cos(100°)}=-5sin(100°)/cos(100°)
x{√3+sin(80°)/cos(80°)}=5sin(80°)/cos(80°)
x{(√3/2)cos(80°)+(1/2)sin(80°)}=5sin(80°)/2
x{sin(60°)cos(80°)+cos(60°)sin(80°)}=5sin(80°)/2
xsin(140°)=5sin(80°)/2
xsin(40°)=5sin(80°)/2
xsin(40°)=5sin(40°)cos(40°)
x=5cos(40°)
y=5√3cos(40°)
∴
D=(5cos(40°),(5√3)cos(40°))

A,Dの内積は
(A,D)
=25cos(40°){cos(80°)+√3sin(80°)}
=50cos(40°){(1/2)cos(80°)+(√3/2)sin(80°)}
=50cos(40°){sin(30°)cos(80°)+cos(30°)sin(80°)}
=50cos(40°)sin(110°)
=50cos(40°)sin(70°)
=25{sin(110°)+sin(30°)}
=25{sin(70°)+1/2}

|BD|の長さの2乗は
|D|^2=25(1+3)cos(40°)^2
|D|^2=100cos(40°)^2
|D|=10cos(40°)
|D|^2=50{1+cos(80°)}

|DA|の長さの2乗は
|A-D|^2
=(A-D,A-D)
=|A|^2-2(A,D)+|D|^2
=25-25{2sin(70°)+1}+50{1+cos(80°)}
=50{1+cos(80°)-sin(70°)}
=50{1+sin(10°)-sin(70°)}
=50{1+sin(10°)+sin(-70°)}
=50{1-2sin(30°)cos(40°)}
=50{1-cos(40°)}
=100{sin(20°)}^2

|DA|の長さは
|A-D|=10sin(20°)

cos∠ADB
=(DA,DB)/(|DB||DA|)
=(A-D,B-D)/(|B-D||A-D|)
=(A-D,-D)/(|D||A-D|)
={|D|^2-(A,D)}/(|D||A-D|)
=|D|/|A-D|-(A,D)/(|D||A-D|)
=cos(40°)/sin(20°)-25{sin(70°)+1/2}/(100cos(40°)sin(20°))
=cos(40°)/sin(20°)-{2sin(70°)+1}/(8cos(40°)sin(20°))
=[8{cos(40°)}^2-2sin(70°)-1]/(8cos(40°)sin(20°))
=[4{1+cos(80°)}-2sin(70°)-1]/(8cos(40°)sin(20°))
=[3+4cos(80°)-2sin(70°)]/(8cos(40°)sin(20°))
=[3+4sin(10°)-2cos(20°)]/[2{√3-2sin(20°)}]
=[3+4sin(10°)-4{(1/2)cos(20°)-(√3/2)sin(20°)+(√3/2)sin(20°)}]/[2{√3-2sin(20°)}]
=[3+4sin(10°)-4{sin(30°)cos(20°)-cos(30°)sin(20°)+(√3/2)sin(20°)}]/[2{√3-2sin(20°)}]
=[3+4sin(10°)-4{sin(10°)+(√3/2)sin(20°)}]/[2{√3-2sin(20°)}]
={3-(2√3)sin(20°)}/[2{√3-2sin(20°)}]
=√3{√3-2sin(20°)}/[2{√3-2sin(20°)}]
=√3/2
∴
∠ADB=30°