+0

0
182
2

There's one triangle satisfying  $$\angle BCA = 60^{\circ}, AB = 12$$ and $$AC = 10$$ shown below:

The only possible value of $$BC$$ is $$a$$, as shown as above. What is $$a$$?

Apr 15, 2022

#1
+14231
+2

What is a?

Hello Guest!

$$c^2=a^2+b^2-2bc\ cos \alpha\\ 12^2=a^2+10^2-2\cdot a\cdot 10\cdot cos\ 60°\\ a^2-10a-44=0\\ a=5\pm \sqrt{25+44}\\ \color{blue}a=13.31$$

!

Apr 15, 2022

#1
+14231
+2

What is a?

Hello Guest!

$$c^2=a^2+b^2-2bc\ cos \alpha\\ 12^2=a^2+10^2-2\cdot a\cdot 10\cdot cos\ 60°\\ a^2-10a-44=0\\ a=5\pm \sqrt{25+44}\\ \color{blue}a=13.31$$

!

asinus Apr 15, 2022
#2
+2532
+1

Here's another way: Draw altitude $$AM$$.

$$\triangle {AMC}$$ is a 30-60-90 triangle, meaning $$CM = 5$$ and $$AM = 5 \sqrt3$$

Now, $$\triangle {AMB}$$ is a right triangle, so applying the Pythagorean Theorem to $$\triangle AMB$$, we find that $$MB = \sqrt{69}$$

Since $$CB = CM + MB$$$$CB = \color{brown}\boxed{5 + \sqrt{69}}$$

Apr 15, 2022