Please consider the attached graph.
We have been given that a helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. The measure of angle ABC is 150°. We are asked to find the area of triangle ABC.
We will use trigonometric area formula to solve our given problem.
, where angle b is angle between sides a and c.
For our given triangle 
and measure of angle b is 150 degrees.
Therefore, the area of the triangle ABC is 10 square kilo-meters and option 'c' is the correct choice.
<span>angle measures in order from greatest to least.
<BAC, <ABC, <ACB
hope it help.</span>
Answer:
(2,1)
Step-by-step explanation:
We can see that the given triangle.
The coordinates of A are (-1,5) .
The coordinates of B are (-1,-3).
The coordinates of C are (5,-3).
Distance formula: 
AB=
units
BC=
units
AC=
units
Pythagoras theorem:
Therefore, 
When a triangle satisfied the Pythagoras theorem then, the triangle is right triangle.
Hence, the given triangle is a right triangle.
We know that circum-center of right triangle is the mid point of hypotenuse.
Mid-point formula:
Using this formula then, we get
Mid-point of hypotenuse AC is given by
Hence, the circum-center of triangle is (2,1).
Answer:
![-7ab\sqrt[3]{3ab^2}](https://tex.z-dn.net/?f=-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D)
Step-by-step explanation:
Remove perfect cubes from under the radical and combine like terms.
![2ab\sqrt[3]{192ab^2}-5\sqrt[3]{81a^4b^5}=2ab\sqrt[3]{4^3\cdot 3ab^2}-5\sqrt[3]{(3ab)^3\cdot 3ab^2}\\\\=(8ab -15ab)\sqrt[3]{3ab^2}=\boxed{-7ab\sqrt[3]{3ab^2} }](https://tex.z-dn.net/?f=2ab%5Csqrt%5B3%5D%7B192ab%5E2%7D-5%5Csqrt%5B3%5D%7B81a%5E4b%5E5%7D%3D2ab%5Csqrt%5B3%5D%7B4%5E3%5Ccdot%203ab%5E2%7D-5%5Csqrt%5B3%5D%7B%283ab%29%5E3%5Ccdot%203ab%5E2%7D%5C%5C%5C%5C%3D%288ab%20-15ab%29%5Csqrt%5B3%5D%7B3ab%5E2%7D%3D%5Cboxed%7B-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D%20%7D)
