Question

( 120^{circ}+20=180^{circ} )
( Rightarrow quad theta=30^{circ} )
( Rightarrow mid a=c )
( cos left(20^{circ}right)=frac{a^{2}+c^{2}-b^{2}}{2 a c}=-frac{1}{2} quad(text { cosine } r w(4) )

# 3. For a triangle shown in the figure, side CA is 10 m, angle ZA and angle ZC are equal then : b=10m (1) Side a = side c = 10m (2) side a # side c (3) side a = side c = 10,3 m (4) side a = side c = tem

Solution