9514 1404 393
Answer:
obtuse
Step-by-step explanation:
The law of cosines tells you ...
b² = a² +c² -2ac·cos(B)
Substituting for a²+c² using the given equation, we have ...
b² = b²·cos(B)² -2ac·cos(B)
We can subtract b² to get a quadratic in standard form for cos(B).
b²·cos(B)² -2ac·cos(B) -b² = 0
Solving this using the quadratic formula gives ...
![\cos(B)=\dfrac{-(-2ac)\pm\sqrt{(-2ac)^2-4(b^2)(-b^2)}}{2b^2}\\\\\cos(B)=\dfrac{ac}{b^2}\pm\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}](https://tex.z-dn.net/?f=%5Ccos%28B%29%3D%5Cdfrac%7B-%28-2ac%29%5Cpm%5Csqrt%7B%28-2ac%29%5E2-4%28b%5E2%29%28-b%5E2%29%7D%7D%7B2b%5E2%7D%5C%5C%5C%5C%5Ccos%28B%29%3D%5Cdfrac%7Bac%7D%7Bb%5E2%7D%5Cpm%5Csqrt%7B%5Cleft%28%5Cdfrac%7Bac%7D%7Bb%5E2%7D%5Cright%29%5E2%2B1%7D)
The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...
![\cos(B)=\dfrac{ac}{b^2}-\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}](https://tex.z-dn.net/?f=%5Ccos%28B%29%3D%5Cdfrac%7Bac%7D%7Bb%5E2%7D-%5Csqrt%7B%5Cleft%28%5Cdfrac%7Bac%7D%7Bb%5E2%7D%5Cright%29%5E2%2B1%7D)
The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.
Answer:
240
Step-by-step explanation:
you do 3x10 which is 30 then multiply that by 8 and you have your answer.
Using the three trigonometric functions can be used to calculate unknown side lengths and angle in a right-angle triangle only.
But, if ever you we're given to find a non right-angle triangle, you will have to use sine rule or cosine rule.
2 6/7 = 2 12/14
1/2 = 7/14
12/14 - 7/14 = 5/14
Answer: 2 5/14
Let me know if you need a better explanation.