Question

exercitiul 8 va rog, am nevoie de el rapid

Answer 1

I

a)

aplic Pitagora in ADB si rezulta ca

DB^2=AB^2-AD^2=2*6^2-6^2=6^2 DB=6

sinB=AD/AB=6/6rad2=rad2/2 rezulta B=45grd

b)

aplic Pitagora in ADC si rezulta ca

AC^2=CD^2+AD^2=12+36=48 AC=6rad3

sinC=AD/AC=6/6rad3=rad3/3 C=arcsin rad3/3

cu teorema sinusurilor

BC/sinA=AC/sinB sin A=BCsinB/AC=[(6+2rad3)*rad2/2]/6rad3

II

cu Pitagora NP^2=MN^2+MP^2=3*5^2+5^2=4*5^2 NP=2*5=10

III

triunghiurile MQN si MQP sunt congruente (MQ-latura comuna, QN=QP si sunt dreptunghice)

rezulte ca MN=MP=4rad5

aplic functia tangenta in triunghiul MQN

tgN=MQ/NQ=0,5=1/2, deci NQ=2MQ

Aplic Pitagora in MQN

MN^2=MQ^2+NQ^2=MQ^2+4MQ^2=5MQ^2

deci 16*5=5*MQ^2

MQ=4 adica NQ=2*4=8 si deci BC=2*NQ=2*8=16

Perimetrul=P=2*4rad5+16=8rad5+16.