а)
∠ B A D = α \angle BAD = \alpha ∠ B A D = α ∠ A B C = 180 – α \angle ABC = 180 – \alpha ∠ A B C = 180– α ∠ C Q P = α \angle CQP = \alpha ∠ C QP = α B P C Q BPCQ B P C Q ∠ P Q N = 180 – α \angle PQN = 180 – \alpha ∠ P QN = 180– α M M M N N N P P P Q Q Q
Проведем C L ∣ ∣ A B ; A B C L CL || AB; ABCL C L ∣∣ A B ; A B C L A B = C L AB = CL A B = C L Δ L C D \Delta LCD Δ L C D L D = 17 – A L = 17 – 4 = 13 ∠ L C D = α ∠ C D L = β LD = 17 – AL = 17–4 = 13\\ ∠LCD = \alpha\\ ∠CDL = \beta L D = 17– A L = 17–4 = 13 ∠ L C D = α ∠ C D L = β cos β = 16 65 ; cos α = 5 13 \cos\beta =\frac{16}{65}; \cos\alpha =\frac{5}{13} cos β = 65 16 ; cos α = 13 5 sin α = 12 13 \sin\alpha =\frac{12}{13} sin α = 13 12 P N PN P N M N MN M N P N = 10 R = P N 2 sin α = 10 2 ⋅ 12 ⋅ 13 = 65 12 M Q 2 sin β = 65 12 sin β = 63 65 M Q = 65 ⋅ sin β 6 = 65 ⋅ 63 65 6 = 21 2 M N = A D + B C 2 = 21 2 ∠ M N Q = M Q N = β ⇒ ∠ Q M N = 180 − 2 β Q N sin ( 180 − 2 β ) = 2 R = 65 6 Q N sin 2 β = 65 6 ⇒ Q N = 65 6 ⋅ sin 2 β = 65 6 ⋅ 2 ⋅ 63 65 ⋅ 16 65 = 336 65 PN=10\\ R=\frac{PN}{2\sin\alpha}=\frac{10}{2\cdot 12}\cdot 13=\frac{65}{12}\\ \frac{MQ}{2\sin\beta}=\frac{65}{12}\\ \frac{\sin\beta =63}{65}\\ MQ=\frac{65\cdot\sin\beta}{6}=\frac{65\cdot\frac{63}{65}}{6}=\frac{21}{2}\\ MN=\frac{AD+BC}{2}=\frac{21}{2}\\ \angle MNQ=MQN=\beta\Rightarrow\angle QMN=180-2\beta\\ \frac{QN}{\sin (180-2\beta)}=2R=\frac{65}{6}\\ \frac{QN}{\sin 2\beta}=\frac{65}{6}\Rightarrow QN=\frac{65}{6}\cdot\sin 2\beta=\frac{65}{6}\cdot 2\cdot\frac{63}{65}\cdot\frac{16}{65}=\frac{336}{65} P N = 10 R = 2 s i n α P N = 2 ⋅ 12 10 ⋅ 13 = 12 65 2 s i n β M Q = 12 65 65 s i n β = 63 M Q = 6 65 ⋅ s i n β = 6 65 ⋅ 65 63 = 2 21 M N = 2 A D + B C = 2 21 ∠ M N Q = M QN = β ⇒ ∠ QM N = 180 − 2 β s i n ( 180 − 2 β ) QN = 2 R = 6 65 s i n 2 β QN = 6 65 ⇒ QN = 6 65 ⋅ sin 2 β = 6 65 ⋅ 2 ⋅ 65 63 ⋅ 65 16 = 65 336 336 65 \frac{336}{65} 65 336
