Answer:
I) ∠CAB measures 60°.
II) BC measures approximately 15.23 cm.
*Please read notes.
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=(1)/(2)ab\sin(C)](https://img.qammunity.org/2022/formulas/mathematics/high-school/si34v87x7ufdb8x6z4c2zs7k1kr0dcfy1b.png)
Where a and b are two side lengths, and C is the angle between the two sides.
Substitute:
![\displaystyle 45=(1)/(2)(10√(3))(6)\sin(A)](https://img.qammunity.org/2022/formulas/mathematics/high-school/zbc3pcm11gqvo9uiqkce9xv70s3fwdwh7v.png)
Simplify:
![\displaystyle 45=30√(3)\sin(A)](https://img.qammunity.org/2022/formulas/mathematics/high-school/pen1xmk28joem4zuqg7426gj6q5ue1ql2e.png)
So:
![\displaystyle \sin(A)=(45)/(30√(3))=(3)/(2\sqrt3)=(√(3))/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/94vwbsfz2unxtbwolrqm0ei2wp9mf4vmxh.png)
Take the inverse sine of both sides. Use a calculator. Thus:
![\displaystyle m\angle A=\sin^(-1)\left((√(3))/(2)\right)=60^\circ](https://img.qammunity.org/2022/formulas/mathematics/high-school/4s0k0jtutbeizj612rtxc0dafo0cdx44m5.png)
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://img.qammunity.org/2022/formulas/mathematics/high-school/56r9mcv55bopi4dozk324h2xkneb5wssj2.png)
Where a and b 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√(3))^2+(6)^2-2(10√(3))(6)\cos(60^\circ)](https://img.qammunity.org/2022/formulas/mathematics/high-school/3ulriknxf1dblfhh5q9swntmeyi1ejroks.png)
Simplify:
![\displaystyle BC^2=336-120√(3)\left((1)/(2)\right)=336-60√(3)](https://img.qammunity.org/2022/formulas/mathematics/high-school/c6bdwv9if1x2ps7eebshw7hbpkm5u9n015.png)
Take the square root of both sides:
![\displaystyle BC=\sqrt{336-60√(3)}\approx 15.23](https://img.qammunity.org/2022/formulas/mathematics/high-school/qp32tjenzfvazsrodmumf16t28msx40v56.png)
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.