Area of a triangle given 2 sides and angle

[lordbodom]

I am doing a problem to find the area of a triangle with the values, AB=8cm, BC= 12 cm and ABC=60°.

The formula to calculate this i know is (1/2)(bc)sinA.

Putting the values, i get:
(1/2)(12.8)sin60
(1/2)48.sin60
24sin60.

Now i know that sine 60={(sqr3)/2}. so the solution would be 24.{(sqr3)/2} or 41.56..

Looking at the answer though, it is 55.42.

What am i missing here?

 

[lookagain]

I am doing a problem to find the area of a triangle with the values, AB=8cm, BC= 12 cm and ABC=60°.

The formula to calculate this i know is (1/2)(bc)sinA.

Putting the values, i get:
(1/2)(12.8)sin60
(1/2)48.sin60
24sin60.

Now i know that sine 60={(sqr3)/2}. so the solution would be 24.{(sqr3)/2} or 41.56..

Looking at the answer though, it is 55.42. <----- The answer is wrong, not including lacking the unit of centimeters.

What am i missing here?

Let's be more careful with the description of the problem:


The length of AB = 8 cm, the length of BC = 12 cm, and the measure of angle ABC = 60°.

Here's another approach to support validating your work:

The area of any triangle is (1/2)bh, b = length of the base, and h = the height

From point A, drop a perpendicular so that it meets BC. Call the place where the perpendicular
meets side BC point D. This makes two right triangles.
Because the measure of angle ABC equals 60°, that means the remaining angle in the right triangle
on the left equals 30°. As this is a 30°-60°-90° triangle, the ratio of the sides is \(\displaystyle 1:\sqrt{3}:2.\)

From this, the height of the triangle is opposite angle ABC and it is \(\displaystyle 4\sqrt{3} \ cm.\)

Side BC can then be the base of the triangle, and its length is 12 cm.


The area

\(\displaystyle = \ \dfrac{1}{2}(12 \ cm)(4\sqrt{3} \ cm) \ square \ \ centimeters\)

\(\displaystyle = \ 24\sqrt{3} \ square \ \ centimeters \)

\(\displaystyle \approx \ 41.57 \ square \ \ centimeters\)

 

[lordbodom]

Let's be more careful with the description of the problem:


The length of AB = 8 cm, the length of BC = 12 cm, and the measure of angle ABC = 60°.

Here's another approach to support validating your work:

The area of any triangle is (1/2)bh, b = length of the base, and h = the height

From point A, drop a perpendicular so that it meets BC. Call the place where the perpendicular
meets side BC point D. This makes two right triangles.
Because the measure of angle ABC equals 60°, that means the remaining angle in the right triangle
on the left equals 30°. As this is a 30°-60°-90° triangle, the ratio of the sides is \(\displaystyle 1:\sqrt{3}:2.\)

From this, the height of the triangle is opposite angle ABC and it is \(\displaystyle 4\sqrt{3} \ cm.\)

Side BC can then be the base of the triangle, and its length is 12 cm.


The area

\(\displaystyle = \ \dfrac{1}{2}(12 \ cm)(4\sqrt{3} \ cm) \ square \ \ centimeters\)

\(\displaystyle = \ 24\sqrt{3} \ square \ \ centimeters \)

\(\displaystyle \approx \ 41.57 \ square \ \ centimeters\)

Thank you for the reply.Sorry, i missed to type sq. centimeters. So if that is the solution, then the solution given i have must be a typo or wrong then?

 

[Mrspi]

Thank you for the reply.Sorry, i missed to type sq. centimeters. So if that is the solution, then the solution given i have must be a typo or wrong then?

Your answer is incorrect because you evaluated (1/2)*(12*8)sin 60 as 24*sin 60....

(1/2)*(12*8) = (1/2)*96, or 48

So, (1/2)*(12*8)sin 60 = 48 sin 60

See if you get an answer now which agrees with the one given in the previous post.

 

[lordbodom]

Your answer is incorrect because you evaluated (1/2)*(12*8)sin 60 as 24*sin 60....

(1/2)*(12*8) = (1/2)*96, or 48

So, (1/2)*(12*8)sin 60 = 48 sin 60

See if you get an answer now which agrees with the one given in the previous post.

I am still not getting the correct answer.

so i get 48 sin 60..

wouldnt that be (48).{(sqr 3)/2}, where you would have the 48 cancel out the 2 giving 24(square root 3)?

 

[Deleted member 4993]

I am still not getting the correct answer.

so i get 48 sin 60..

wouldnt that be (48).{(sqr 3)/2}, where you would have the 48 cancel out the 2 giving 24(square root 3)?

If you posted the correct problem:

find the area of a triangle with the values, AB=8cm, BC= 12 cm and ABC=60°.

Then

The area of the triangle = 24√3 = 41.569 cm²

 

[lookagain]

I am still not getting the correct answer.

so i get 48 sin 60..

wouldnt that be (48).{(sqr 3)/2}, where you would have the 48 cancel out the 2 giving > > > 24(square root 3) < < < ?

\(\displaystyle \ \ \ \) ? ? ? \(\displaystyle \ \ \)lordbodom, did you not read the last few lines of my post (post # 2)!? Look at the second to the last line in post #2 and look at what you typed in the highlighted part of the quote box immediately above this. It's agreement.

 

[lordbodom]

\(\displaystyle \ \ \ \) ? ? ? \(\displaystyle \ \ \)lordbodom, did you not read the last few lines of my post (post # 2)!? Look at the second to the last line in post #2 and look at what you typed in the highlighted part of the quote box immediately above this. It's agreement.

Thanks. I misread that as I was making an error. For the 5.7, did you just round up the 5.69 I am getting?