In triangle ABC, AB=2 square root 2, BC=8,angle ABC=45 degrees.Find area ABC.

a. 4 square 2
b. 8
c. 8 square 2
d. 16
e. 16 square 2

please answer and explain

5 answers

Use the law of cosines to find AC:
8+64-2*2√2*8/√2
72 - 32
AC = 2√10

Now you can use Heron's formula to get the area = 8
what is heron's formula

I am grade 7
Hmm. So, you haven't had calculus, trig, or much algebra. What have you studied to enable you to find the area of a triangle, given two sides and the included angle?

What chapter of the text presents this problem? I'd love to help you solve it, but some context would be nice.
OK. I have an idea, if you know the Pythagorean Theorem. Draw ABC with the base BC, and point A at the top.

Now, the line AB is length 2√2 and angle B is 45°. Draw the altitude AD to the base BC.

Now you have a right triangle BDA with hypotenuse 2√2. Its legs AD and BD are both of length 2, since BD^2+AD^2 = 8 and BD=AD. So, 2AD^2 = 8, AD^2 = 4, AD=2.

So, for triangle ABC, the base BC=8 and the altitude DA=2, so the area is 8.
Cosec 31° -sec59°