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МатематикаВопрос:
30. Find the missing angles, given that NS = PS = KS.
Ответ:
Since NS = PS, triangle NPS is an isosceles triangle. Therefore, \(\angle PNS = \angle NPS = 25^\circ\).
So, \(\angle NSP = 180^\circ - (25^\circ + 25^\circ) = 180^\circ - 50^\circ = 130^\circ\).
Since NS = KS, triangle NKS is an isosceles triangle. Therefore, \(\angle KNS = \angle NKS = 35^\circ\).
So, \(\angle NSK = 180^\circ - (35^\circ + 35^\circ) = 180^\circ - 70^\circ = 110^\circ\).
\(\angle PSK = 360^\circ - (130^\circ + 110^\circ) = 360^\circ - 240^\circ = 120^\circ\).
Since PS = KS, triangle PKS is an isosceles triangle. \(\angle SPK = \angle SKP\). Therefore, \(\angle SPK + \angle SKP + \angle PSK = 180^\circ\).
\(2 \angle SPK + 120^\circ = 180^\circ\).
\(2 \angle SPK = 60^\circ\).
\(\angle SPK = 30^\circ\) and \(\angle SKP = 30^\circ\).
**Answer: \(\angle NPS = 25^\circ, \angle NSP = 130^\circ, \angle NKS = 35^\circ, \angle NSK = 110^\circ, \angle SPK = 30^\circ, \angle SKP = 30^\circ\)**
So, \(\angle NSP = 180^\circ - (25^\circ + 25^\circ) = 180^\circ - 50^\circ = 130^\circ\).
Since NS = KS, triangle NKS is an isosceles triangle. Therefore, \(\angle KNS = \angle NKS = 35^\circ\).
So, \(\angle NSK = 180^\circ - (35^\circ + 35^\circ) = 180^\circ - 70^\circ = 110^\circ\).
\(\angle PSK = 360^\circ - (130^\circ + 110^\circ) = 360^\circ - 240^\circ = 120^\circ\).
Since PS = KS, triangle PKS is an isosceles triangle. \(\angle SPK = \angle SKP\). Therefore, \(\angle SPK + \angle SKP + \angle PSK = 180^\circ\).
\(2 \angle SPK + 120^\circ = 180^\circ\).
\(2 \angle SPK = 60^\circ\).
\(\angle SPK = 30^\circ\) and \(\angle SKP = 30^\circ\).
**Answer: \(\angle NPS = 25^\circ, \angle NSP = 130^\circ, \angle NKS = 35^\circ, \angle NSK = 110^\circ, \angle SPK = 30^\circ, \angle SKP = 30^\circ\)**
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