Question

Fie trigunghiul mnp isoscel cul varful in m are un unghi de 120grade si pm=38 cm. Calculati doferenta de la P la inaltimra dusa din N

Answer 1

notam triunghiul PMN
ducem ND_I_PM (si deoarece este obtuzunghic inaltimea va cadea in afara laturii MP
vom nota MD=x
m(PMN)=120⁰
ΔPMN fiind isoscel ⇒ m(<MPN)=m(<MNP)=(180⁰-120⁰)/2=30⁰
sin 30⁰=1/2   DN/PN=1/2   ⇒ PN=2DN
m(<DMN)=180⁰-120⁰=60⁰
tg 60⁰=√3     DN/x=√3   ⇒ DN=x√3
PN²=(38+x)²+DN²
(2DN) ²=(38+x)²+DN²     dar  DN=x√3
(2x√3)²=1444+76x+x²+(x√3)²
12x²=1444+76x+x²+3x²
8x²-76x-1444=0
4x²-38x-722=0
Δ=38²+4*722*4=1444+11552=12996
x=(38+√12996)/2*4=(38+114)/8=152/8=19
MD=19 cm
PD=38+19=97 cm