Question

please tell me the right answer I want urgent​

Answer 1

Sum of all angles of a triangle= 180°

a) R= 95°

Q= 45°

Let P be x°

Therefore, 95°+45°+x°=180°

140+x°=180°

x°=180°-140°

x°=40°

So, R= 95°

Q= 45°

P= 40°

P= 40°Obtuse Triangle

P= 40°Obtuse Triangle

b) A= 65°

B= 35°

Let C be x°

Therefore, 65°+35°+x°=180°

100+x°=180°

x°=180°-100°

x°=80°

So, A= 65°

B= 35°

C= 80°

= 80°Acute Triangle

c) A= 80°

B= 80°

Let C be x°

Therefore, 80°+80°+x°=180°

160+x°=180°

x°=180°-160°

x°=20°

So, A= 80°

B= 80°

C= 20°

C= 20°Acute Triangle

d) P= 140°

Let Q and R be x°

Therefore, 140°+ x +x°=180°

140 + 2x°=180°°

2x°=40

x°=20°

So, P= 140°

Q= 20°

R= 20°

R= 20°Obtuse Triangle

e) B= 80°

B - C= 40°

80 - 40°= C

40°= C

Let A be x°

Therefore, 80°+ 40°+ x°=180°

120° + x°=180°

x°=(180-120)°

x°=60°

So, A= 60°

B= 80°

C= 40°

C= 40°Acute Triangle