Solve ABC if C=90 degrees, B=20 degrees, and b=10. Round measures of side to the nearest tenth and measures of angles to the nearest degree.
So all I need to do for this problem is find A,a,c right? I know that A=70 b/c 90+20=110 and 180-110=70 but how do I find a,c?
To find the lengths of side a and side c in triangle ABC, you can use the Law of Sines, which states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant.
In this case, you already know the measures of angles A (70 degrees), B (20 degrees), and C (90 degrees), and the length of side b (10 units). So you can use the Law of Sines to find the lengths of sides a and c.
The Law of Sines can be written as:
a / sin(A) = b / sin(B) = c / sin(C)
Since you want to find sides a and c, substitute the values you know into the equation:
a / sin(70) = 10 / sin(20)
Now you can rearrange the equation to solve for a:
a = (10 * sin(70)) / sin(20)
Use a calculator to find the approximate value of sin(70) and sin(20), and then substitute those values into the equation:
a ≈ (10 * 0.9397) / 0.3420
a ≈ 27.46
So side a is approximately 27.5 units.
To find side c, you can use the Law of Sines again, but this time with angle C:
c / sin(C) = 10 / sin(20)
Rearranging the equation:
c = (10 * sin(C)) / sin(20)
Since sin(90) = 1, the equation simplifies to:
c = 10 / sin(20)
Using a calculator to find the approximate value of sin(20):
c ≈ 10 / 0.3420
c ≈ 29.24
So side c is approximately 29.2 units.
Therefore, in triangle ABC with C = 90 degrees, B = 20 degrees, and b = 10 units, side a is approximately 27.5 units and side c is approximately 29.2 units.
You have a right-angled triangle, so basic trig can be used.
tan20º = 10/a and sin20º = 10/c, so
a = 10/tan20 and c=10/sin20