Answer on Question#43375 – Math – Geometry
Question
A small triangle abc lie in a large triangle ABC with same base and same top point i.e., a and A is same point. If < A B C = 40 <\mathrm{ABC}=40 < ABC = 40 < A C B = 35 <\mathrm{ACB}=35 < ACB = 35 < a b c = p <\mathrm{abc}=p < abc = p < a c b = q <\mathrm{acb}=q < acb = q < b a c = 32 <\mathrm{bac}=32 < bac = 32 < B A b = s <\mathrm{BAb}=s < BAb = s < C A c = r <\mathrm{CAc}=r < CAc = r p , q , r , p, q, r, p , q , r , s s s
Solution
In Δ A B C \Delta ABC Δ A BC
< B A C + < A B C + < A C B = 180 ∘ , whence < B A C = 180 ∘ − < A B C − < A C B = 180 ∘ − 40 ∘ − 35 ∘ = 105 ∘ < B A C = < B A b + < b A c + < c A C = s + 32 ∘ + r = 105 ∘ s + r = 73 ∘ \begin{array}{l}
<\mathrm{BAC} + <\mathrm{ABC} + <\mathrm{ACB} = 180{}^{\circ}, \text{whence} \\
<\mathrm{BAC} = 180{}^{\circ} - <\mathrm{ABC} - <\mathrm{ACB} = 180{}^{\circ} - 40{}^{\circ} - 35{}^{\circ} = 105{}^{\circ} \\
<\mathrm{BAC} = <\mathrm{BAb} + <\mathrm{bAc} + <\mathrm{cAC} = s + 32{}^{\circ} + r = 105{}^{\circ} \\
s + r = 73{}^{\circ} \\
\end{array} < BAC + < ABC + < ACB = 180 ∘ , whence < BAC = 180 ∘ − < ABC − < ACB = 180 ∘ − 40 ∘ − 35 ∘ = 105 ∘ < BAC =< BAb + < bAc + < cAC = s + 32 ∘ + r = 105 ∘ s + r = 73 ∘
In Δ a b c \Delta abc Δ ab c
< b a c + < a c b + < c b a = 180 ∘ , whence < a c b + < c b a = p + q = 180 ∘ − < b a c = 180 ∘ − 32 ∘ = 148 ∘ \begin{array}{l}
<\mathrm{bac} + <\mathrm{acb} + <\mathrm{cba} = 180{}^{\circ}, \text{whence} \\
<\mathrm{acb} + <\mathrm{cba} = p + q = 180{}^{\circ} - <\mathrm{bac} = 180{}^{\circ} - 32{}^{\circ} = 148{}^{\circ} \\
\end{array} < bac + < acb + < cba = 180 ∘ , whence < acb + < cba = p + q = 180 ∘ − < bac = 180 ∘ − 32 ∘ = 148 ∘
In Δ B b a \Delta Bba Δ B ba
< A B b + < B b a + < B A b = 180 ∘ , hence < B b a = 180 ∘ − 40 ∘ − s = 140 ∘ − s < B b a + < a b c = 180 ∘ , i.e. ( 140 ∘ − s ) + p = 180 ∘ , whence p − s = 40 ∘ \begin{array}{l}
<\mathrm{ABb} + <\mathrm{Bba} + <\mathrm{BAb} = 180{}^{\circ}, \text{hence} <\mathrm{Bba} = 180{}^{\circ} - 40{}^{\circ} - s = 140{}^{\circ} - s \\
<\mathrm{Bba} + <\mathrm{abc} = 180{}^{\circ}, \text{i.e. } (140{}^{\circ} - s) + p = 180{}^{\circ}, \text{whence } p - s = 40{}^{\circ} \\
\end{array} < ABb + < Bba + < BAb = 180 ∘ , hence < Bba = 180 ∘ − 40 ∘ − s = 140 ∘ − s < Bba + < abc = 180 ∘ , i.e. ( 140 ∘ − s ) + p = 180 ∘ , whence p − s = 40 ∘
In Δ C c a \Delta Cca Δ C c a
< A C c + < C c a + < C a c = 180 ∘ , hence < C c a = 180 ∘ − 35 ∘ − r = 145 ∘ − r < C c a + < a c b = 180 ∘ , i.e. ( 145 ∘ − r ) + q = 180 ∘ , whence q − r = 35 ∘ \begin{array}{l}
<\mathrm{ACc} + <\mathrm{Cca} + <\mathrm{Cac} = 180{}^{\circ}, \text{hence} <\mathrm{Cca} = 180{}^{\circ} - 35{}^{\circ} - r = 145{}^{\circ} - r \\
<\mathrm{Cca} + <\mathrm{acb} = 180{}^{\circ}, \text{i.e. } (145{}^{\circ} - r) + q = 180{}^{\circ}, \text{whence } q - r = 35{}^{\circ} \\
\end{array} < ACc + < Cca + < Cac = 180 ∘ , hence < Cca = 180 ∘ − 35 ∘ − r = 145 ∘ − r < Cca + < acb = 180 ∘ , i.e. ( 145 ∘ − r ) + q = 180 ∘ , whence q − r = 35 ∘
In Δ A b c \Delta Abc Δ A b c
< A C b + < C b a + < b A C = 35 ∘ + p + ( 32 ∘ + r ) = 180 ∘ , whence p + r = 113 ∘ \begin{array}{l}
<\mathrm{ACb} + <\mathrm{Cba} + <\mathrm{bAC} = 35{}^{\circ} + p + (32{}^{\circ} + r) = 180{}^{\circ}, \text{whence} \\
p + r = 113{}^{\circ} \\
\end{array} < ACb + < Cba + < bAC = 35 ∘ + p + ( 32 ∘ + r ) = 180 ∘ , whence p + r = 113 ∘
In Δ A B C \Delta ABC Δ A BC
< A B C + < B A C + < A C B = 40 ∘ + ( s + 32 ∘ ) + q = 180 ∘ , whence q + s = 108 ∘ \begin{array}{l}
<\mathrm{ABC} + <\mathrm{BAC} + <\mathrm{ACB} = 40{}^{\circ} + (s + 32{}^{\circ}) + q = 180{}^{\circ}, \text{whence} \\
q + s = 108{}^{\circ} \\
\end{array} < ABC + < BAC + < ACB = 40 ∘ + ( s + 32 ∘ ) + q = 180 ∘ , whence q + s = 108 ∘
Thus, we have the set of 6 linear equations with 4 unknown values:
{ s + r = 73 ∘ p + q = 148 ∘ p − s = 40 ∘ q − r = 35 ∘ q + s = 108 ∘ p + r = 113 ∘ \left\{
\begin{array}{l}
s + r = 73{}^{\circ} \\
p + q = 148{}^{\circ} \\
p - s = 40{}^{\circ} \\
q - r = 35{}^{\circ} \\
q + s = 108{}^{\circ} \\
p + r = 113{}^{\circ} \\
\end{array}
\right. ⎩ ⎨ ⎧ s + r = 73 ∘ p + q = 148 ∘ p − s = 40 ∘ q − r = 35 ∘ q + s = 108 ∘ p + r = 113 ∘
There is not the only solution of the set of equations.
Many solutions satisfying the condition may be found. For example
p = 53 ∘ , q = 95 ∘ , r = 60 ∘ , s = 13 ∘ or p = 57 ∘ , q = 91 ∘ , r = 56 ∘ , s = 17 ∘ or p = 90 ∘ , q = 58 ∘ , r = 23 ∘ , s = 50 ∘ \begin{array}{l}
p = 53{}^{\circ}, q = 95{}^{\circ}, r = 60{}^{\circ}, s = 13{}^{\circ} \text{ or} \\
p = 57{}^{\circ}, q = 91{}^{\circ}, r = 56{}^{\circ}, s = 17{}^{\circ} \text{ or} \\
p = 90{}^{\circ}, q = 58{}^{\circ}, r = 23{}^{\circ}, s = 50{}^{\circ} \\
\end{array} p = 53 ∘ , q = 95 ∘ , r = 60 ∘ , s = 13 ∘ or p = 57 ∘ , q = 91 ∘ , r = 56 ∘ , s = 17 ∘ or p = 90 ∘ , q = 58 ∘ , r = 23 ∘ , s = 50 ∘
and so on.
Any of such solutions may be the answer.
Answer: p = 53 ∘ , q = 95 ∘ , r = 60 ∘ , s = 13 ∘ p = 53{}^{\circ}, q = 95{}^{\circ}, r = 60{}^{\circ}, s = 13{}^{\circ} p = 53 ∘ , q = 95 ∘ , r = 60 ∘ , s = 13 ∘
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