the triangle ABC., BC = 12cm, ABC = 80 degree and ACB = 30 degree
Calculate the area of triangle ABC.
asked by Kd
today at 3:06pm
angle BAC = 180-80-30 = ....
now find AB using the Sine Law.
Area of triangle = (1/2)(AB)(BC)sin80° = ....
you do the button-pushing.
Let me know what you get.
Is that correct
sin70/a = sin80/12
No, you are not matching them up correctly
should be :
sine(angle1)/side opposite that angle1 = sin(angle2)/side opposite angle2
so sin30/AB = sin70/12
continue ....
Yes, it looks like you're on the right track.
To find angle BAC, you correctly subtracted the given angles ABC and ACB from 180 degrees. So the measure of angle BAC is 180 - 80 - 30 = 70 degrees.
Next, you can use the Law of Sines to find the length of side AB. The Law of Sines states that in any triangle, the ratio of a side length to the sine of its opposite angle is constant. In this case, we can write:
sin(BAC)/AB = sin(ACB)/BC
Plugging in the values, we have:
sin(70)/AB = sin(30)/12
Now, you can solve for AB by cross-multiplying and simplifying the equation:
sin(70) * 12 = AB * sin(30)
AB = (sin(70) * 12) / sin(30)
Now, you can substitute the value of AB into the formula for the area of the triangle:
Area of triangle ABC = (1/2) * AB * BC * sin(80)
Plugging in the values, you have:
Area of triangle ABC = (1/2) * [(sin(70) * 12) / sin(30)] * 12 * sin(80)
Simplifying this expression will give you the area of triangle ABC.