If the vertices of Delta ABC are (a, 0, 0),(0 | vedclass

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(A) Let the vertices be $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.To find $\angle B$,we need the angle between vectors $\vec{BA}$ and $\vec{BC}$.$\ve

Vedclass · Accepted Answer

$\cos^{-1} \frac{b^2}{\sqrt{(a^2 + b^2)(b^2 + c^2)}}$

Vedclass · Answer

(A) Let the vertices be $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.To find $\angle B$,we need the angle between vectors $\vec{BA}$ and $\vec{BC}$.$\vec{BA} = (a - 0, 0 - b, 0 - 0) = (a, -b, 0)$.$\vec{BC} = (0 - 0, 0 - b, c - 0) = (0, -b, c)$.The cosine of the angle $	heta$ at vertex $B$ is given by:$\cos B = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|}$$\vec{BA} \cdot \vec{BC} = (a)(0) + (-b)(-b) + (0)(c) = b^2$.$|\vec{BA}| = \sqrt{a^2 + (-b)^2 + 0^2} = \sqrt{a^2 + b^2}$.$|\vec{BC}| = \sqrt{0^2 + (-b)^2 + c^2} = \sqrt{b^2 + c^2}$.Therefore,$\cos B = \frac{b^2}{\sqrt{a^2 + b^2} \sqrt{b^2 + c^2}}$.Thus,$B = \cos^{-1} \left( \frac{b^2}{\sqrt{(a^2 + b^2)(b^2 + c^2)}} ight)$.

If the vertices of $\Delta ABC$ are $(a, 0, 0)$,$(0, b, 0)$,and $(0, 0, c)$ respectively,then $\angle B = \dots$

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