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
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°
To find the area of triangle ABC, we can use the formula for the area of a triangle:
Area = 1/2 * base * height.
First, let's find the height of the triangle. We know that BC = 8 and angle ABC = 45 degrees. The base of the triangle is BC, and the height is perpendicular to the base and passes through point A.
Since angle ABC is 45 degrees, we can deduce that triangle ABC is an isosceles right triangle because two angles are the same and one of them is 90 degrees.
Since AB = 2√2 and angle ABC is 45 degrees, we can determine that the height is equal to AB. Therefore, the height of the triangle is 2√2.
Now, we can substitute the values of the base and height into the formula for the area of a triangle:
Area = 1/2 * 8 * (2√2)
Simplifying, we get:
Area = 4 * √2
The correct answer is (a) 4√2.