CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In a triangle $$ABC, 2ac \sin \dfrac {1}{2} (A - B + C) =$$


A
a2+b2c2
loader
B
c2+a2b2
loader
C
b2c2a2
loader
D
c2a2b2
loader

Solution

The correct option is D $$c^{2} + a^{2} - b^{2}$$
We know that,

$$A+B+C=\pi$$

$$\Rightarrow A+C=\pi -B$$

$$\Rightarrow A+C-B=A+B+C-2B=2\left( \dfrac { \pi  }{ 2 } -B \right) $$

So, here $$2ac\sin { \dfrac { 1 }{ 2 }  } \left( A+C-B \right)$$ will be

$$2ac\sin \left(  \dfrac{\pi}{2}-B \right)=2ac\cos B$$

By cosine rule, we have

$$2ac\cos B=c^2+a^2-b^2$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image