Question

I need help on khan academy and I’m willing to pay half right when you start the work and when you finish all the work I’ll pay the other half. I am desperate if I don’t pass this class it’s bad for me. Comment your number,snap anything. If not can you please help me with this one problem please thank youu

Answer 1

Answer:

$x\approx 50^\circ$

Step-by-step explanation:

$c^2 = a^2 + b^2 - 2ab(cos(C))$

See the figure below to get the values as:

$7^2=7^2+9^2-2\left(7\right)\left(9\right)cos\left(x\right)\\\\cos(x)=\frac{7^2+9^2-7^2}{2\cdot \:7\cdot \:9}\\\\x\approx 50^\circ$

There are multiple concepts to solve this problem. This is one of the concept used in high school. Other concept to solve this problem is to use the concept of isosceles triangle. An isosceles triangle is a triangle with (at least) two equal sides. The angles shared by the two equal sides are also equal. So that the sum of all the three angles will add up to 180.

$x+x+80=180\\\\2x=100\\\\x=50^{\circ}$

Best Regards!