Given: △abc, m∠a=60° m∠c=45°, ab=8 find: perimeter of △abc, area of △abc pls

Let ∆ABC is a triangle such that side length of AB is c , BC is a and CA is b . 
Given, AB = c = 8 
m∠A=60°
m∠C=45°
m∠B = (180 - 45 - 60)° = 75°

use sine rule to get b and c 
\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}asinA​=bsinB​=csinC​ 

so, \frac{sin60^{\circ}}{8}=\frac{sin45^{\circ}}{c}=\frac{sin75^{\circ}}{b}8sin60∘​=csin45∘​=bsin75∘​ 

sin60°/8 = sin45°/c 

(√3/2)/8 = (1/√2)/c 

√3/16 = 1/√2c => c = 16/√6 

also, sin60°/8 = sin75°/b 

b = 8sin75°/sin60° 

= {8 × (√3 + 1)/2√2}/{√3/2}

= 4√2(√3 + 1)/√3 

hence, perimeter of ∆ABC = a + b + c 
= 8 + 16/√6 + 4√2(√3 + 1)/√3 
= 8 + (16 + 8√3 + 8)/√6 
= 8 + (24 + 8√3)/√6 
= 8 + 4√6 + 4√2 

area of ∆ABC = 1/2 absinC 
= 1/2 × 8 × 4√2(√3 + 1)/√3 × sin45°
= 4 × 4√2(√3 + 1)/√3 × 1/√2 
= 16(√3 + 1)/√3 
= (48 + 16√3)/3 
= 16 + 16/√3 

Dbc if u do the math and stuff that what i got let me know if i was wrong