Question

Find the area of the triangle ABC given angle B=45 degrees, side a = 10, b = sqrt2

Answer 1

step 1) use the law of sines to find angle A
sin(A)/a = sin(B)/b
sin(A)/10 = sin(45)/sqrt(2)
solve for A

step 2) use angle A and B to find angle C
A+B+C = 180 degrees
C = 180-A-B

step 3) use sides 'a' and 'b', along with angle C, in the area formula
area = (1/2)*a*b*sin(C)

Answer 2

Answer choices are

5
5sqrt2
10
10sqrt2

Answer 3

Thanks a bunch! :)

Answer 4

hmm there's probably a typo in the problem
when I tried to solve
sin(A)/10 = sin(45)/sqrt(2)
I ended up with

sin(A) = 5

but that has no solutions

Answer 5

Yeah that's what I got too

Answer 6

Is there a way to find the third side to calculate the area?

Answer 7

law of cosines

c^2 = a^2 + b^2 - 2*a*b*cos(C)

but you'd need to know the angle C

Answer 8

Hmm them so this is unsolvable the way it is?

Answer 9

*then

Answer 10

I just used geogebra to try to construct this triangle visually. I'm seeing that it's not possible with these sides and angle

can I see a screenshot of the full problem?

Answer 11

Yeah I'm using a mobile device though could I send it to you another way?

Answer 12

sure, send it however you want

Answer 13

Sent it

Answer 14

we know 2 sides and the angle between them

so we use the SAS triangle area formula

area = (1/2)*(side1)*(side2)*sin(angle between sides)
area = (1/2)*(a)*(c)*sin(B)
area = (1/2)*(10)*(sqrt(2))*sin(45)

I'll let you finish up

Answer 15

let me know if you have any questions @DiamondAnon2