5) ∠Q=∠M=∠N=180°:3=60° все стороны равны- Δ равносторонний и у него все углы равны по теореме о сумме трёх углов Δ
∠Q=∠M=∠N=180°:3=60°
6)∠E=90°;
∠P=90°-60°=30° по теореме о сумме острых углов прямоугольногоΔ.
7) MD=DN, ΔMDN- равносторонний,∠M и∠N- углы при основанииΔ
∠M=∠N=(180°-100°)/2=40°.
9) MN=NK, ΔMNK - равносторонний ∠M и∠K - углы при основанииΔ
∠M=180°-130°=50°; как смежный с внешним∠
∠M=∠K=50°;∠N=130°-∠K=80°.( как сумма двух углов против внешнего угла треугольника)
10)∠E=180°-140°=40°; как смежный с ∠CEF
∠D=180°-80°-40°=60° ( по теореме о сумме трёх углов).
11)∠C=90, ∠A=180°-150°=30°; ∠B=90-30°=60° по теореме о сумме острых углов прямоугольногоΔ.
а) 0,5:1
б) 100:126:169
(3+4+8)=180°; a=180/15=12
ZA=3*12=36°
ZB=12*4=48°
ZC=12*8=96
a=100
b=126
c=169
Объяснение:
︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎
5) ∠Q=∠M=∠N=180°:3=60° все стороны равны- Δ равносторонний и у него все углы равны по теореме о сумме трёх углов Δ
∠Q=∠M=∠N=180°:3=60°
6)∠E=90°;
∠P=90°-60°=30° по теореме о сумме острых углов прямоугольногоΔ.
7) MD=DN, ΔMDN- равносторонний,∠M и∠N- углы при основанииΔ
∠M=∠N=(180°-100°)/2=40°.
9) MN=NK, ΔMNK - равносторонний ∠M и∠K - углы при основанииΔ
∠M=180°-130°=50°; как смежный с внешним∠
∠M=∠K=50°;∠N=130°-∠K=80°.( как сумма двух углов против внешнего угла треугольника)
10)∠E=180°-140°=40°; как смежный с ∠CEF
∠D=180°-80°-40°=60° ( по теореме о сумме трёх углов).
11)∠C=90, ∠A=180°-150°=30°; ∠B=90-30°=60° по теореме о сумме острых углов прямоугольногоΔ.
а) 0,5:1
б) 100:126:169
(3+4+8)=180°; a=180/15=12
ZA=3*12=36°
ZB=12*4=48°
ZC=12*8=96
a=100
b=126
c=169
Объяснение:
︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎ ︎