Given: △ABC, m∠A=60°,
m∠C=45°, AB=9
Find: Perimeter of △ABC,
Area of △ABC
Pleas be sure to answer BOTH PARTS ... thank you!

Answers

Answer 1

Answer:

p = 32.317

A = 47.911

Step-by-step explanation:

given ∠A = 60

∠C = 45

AB = 9

From sum of all angles in triangle is 180

∠A + ∠B + ∠C = 180

60 + ∠B + 45 = 180

∠B = 180 - 105 = 75

∠B = 75

from sine rule in ΔABC

AB / sin C = BC / sin A

=> 9 / sin 45 = BC / sin 60

=> 9 / 1/[tex]\sqrt{2}[/tex] = BC / [tex]\sqrt{3} / 2[/tex]

=> [tex]9\sqrt{2} = BC * 2 / \sqrt{3}[/tex]

BC = [tex]9\sqrt{2} * \sqrt{3} / 2\\[/tex]

BC = 11.023

length of AC is

AC / sin B = AB / sin C

AC / sin 75 = 9 / sin 45

AC = 9 / 1/[tex]\sqrt{2} * sin 75[/tex]

= [tex]9\sqrt{2} * sin 75\\12.294[/tex]

perimeter of ΔABC

AB + BC + CA

= 9 + 11.023 + 12.294

= 32.317

Area of ΔABC is

= 1/2 * AB * AC * sin A

= 1/2 * 9 * 12.294 * sin 60

= 1/2 * 9 * 12.294 * √3/2

= 47.911


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Answers

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