Answer:
I) ∠CAB measures 60°.
II) BC measures approximately 15.23 cm.
*Please read notes.
Step-by-step explanation:
We are given ΔABC, where AB measures 10√3 cm, AC measures 6 cm, and the triangle has an area of 45 square cm.
And we want to find I) the measure of ∠CAB and II) the length of BC.
I) First, we should always draw a representative triangle so we can determine the sides and angles. This is shown below.
Note that ∠CAB (or simply ∠A) is the angle between the two sides. Thus, we can find the angle by using the alternative formula for the area of a triangle:
![\displaystyle A=\frac{1}{2}ab\sin(C)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%3D%5Cfrac%7B1%7D%7B2%7Dab%5Csin%28C%29)
Where <em>a</em> and <em>b</em> are two side lengths, and C is the angle between the two sides.
Substitute:
![\displaystyle 45=\frac{1}{2}(10\sqrt{3})(6)\sin(A)](https://tex.z-dn.net/?f=%5Cdisplaystyle%2045%3D%5Cfrac%7B1%7D%7B2%7D%2810%5Csqrt%7B3%7D%29%286%29%5Csin%28A%29)
Simplify:
![\displaystyle 45=30\sqrt{3}\sin(A)](https://tex.z-dn.net/?f=%5Cdisplaystyle%2045%3D30%5Csqrt%7B3%7D%5Csin%28A%29)
So:
![\displaystyle \sin(A)=\frac{45}{30\sqrt{3}}=\frac{3}{2\sqrt3}=\frac{\sqrt{3}}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csin%28A%29%3D%5Cfrac%7B45%7D%7B30%5Csqrt%7B3%7D%7D%3D%5Cfrac%7B3%7D%7B2%5Csqrt3%7D%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D)
Take the inverse sine of both sides. Use a calculator. Thus:
![\displaystyle m\angle A=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^\circ](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%5Cangle%20A%3D%5Csin%5E%7B-1%7D%5Cleft%28%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Cright%29%3D60%5E%5Ccirc)
II) To find BC, we will use the Law of Cosines. We do not know whether or not ΔABC is a right triangle, so we cannot use right triangle trigonometry or special right triangles.
The law of cosines is:
![\displaystyle c^2=a^2+b^2-2ab\cos(C)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20c%5E2%3Da%5E2%2Bb%5E2-2ab%5Ccos%28C%29)
Where <em>a</em> and <em>b</em> are side lengths and C is the angle between the side lengths. c is the side length opposite to the angle.
BC is opposite to A. Substitute:
![(BC)^2=(10\sqrt{3})^2+(6)^2-2(10\sqrt{3})(6)\cos(60^\circ)](https://tex.z-dn.net/?f=%28BC%29%5E2%3D%2810%5Csqrt%7B3%7D%29%5E2%2B%286%29%5E2-2%2810%5Csqrt%7B3%7D%29%286%29%5Ccos%2860%5E%5Ccirc%29)
Simplify:
![\displaystyle BC^2=336-120\sqrt{3}\left(\frac{1}{2}\right)=336-60\sqrt{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20BC%5E2%3D336-120%5Csqrt%7B3%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%3D336-60%5Csqrt%7B3%7D)
Take the square root of both sides:
![\displaystyle BC=\sqrt{336-60\sqrt{3}}\approx 15.23](https://tex.z-dn.net/?f=%5Cdisplaystyle%20BC%3D%5Csqrt%7B336-60%5Csqrt%7B3%7D%7D%5Capprox%2015.23)
BC measures about 15.23 cm.
Notes:
In this case, the inverse sine of √3/2 will yield two answers: 60° and 120°. This is an example of an ambiguous case. Both of these angles will work. However, BC will be different. If ∠A is 60°, then BC is about 15.23. However, if ∠A is 120°, then BC is about 20.97. Both will result in the triangle having an area of 45 square cm, as well as AB measuring 10√3 and AC measuring 6.