Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0
1 answer:
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The correct question in the attached figure
we know that
applying the law of sines
g/sinG=k/sinK
g=3 units
G=30°
k=4 units
K=?
so
g/sinG=k/sinK------> g*sinK=k*sinG-----> sinK=k*sinG/g
sinK=4*sin 30°/3-----> 2/3
K=arc sin(2/3)-------> K=41.81°
case 1)
for angle K=41.81°
the sum of the angles (H +G+K)=180°
H=180-(G+K)------> 180-(30+41.81)--------> H=108.19°
h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 108.19°/sin 30°
h=5.70 units
case 2)
for angle K=180°-41.81°---------> K=138.19°
the sum of the angles (H +G+K)=180°
H=180-(G+K)------> 180-(30+138.19)--------> H=11.81°
h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 11.81°/sin 30°
h=1.2 units
therefore
the answer is the option
<span>A. 1.2 or 5.7</span>
we know that
applying the law of sines
g/sinG=k/sinK
g=3 units
G=30°
k=4 units
K=?
so
g/sinG=k/sinK------> g*sinK=k*sinG-----> sinK=k*sinG/g
sinK=4*sin 30°/3-----> 2/3
K=arc sin(2/3)-------> K=41.81°
case 1)
for angle K=41.81°
the sum of the angles (H +G+K)=180°
H=180-(G+K)------> 180-(30+41.81)--------> H=108.19°
h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 108.19°/sin 30°
h=5.70 units
case 2)
for angle K=180°-41.81°---------> K=138.19°
the sum of the angles (H +G+K)=180°
H=180-(G+K)------> 180-(30+138.19)--------> H=11.81°
h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 11.81°/sin 30°
h=1.2 units
therefore
the answer is the option
<span>A. 1.2 or 5.7</span>