If |a| + |b| = |c| and a + b = c,then the ang | vedclass

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(C) Given $a + b = c$. Squaring both sides,we get $|a + b|^2 = |c|^2$. This implies $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$. Also,we are given $|a| + |

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(C) Given $a + b = c$. Squaring both sides,we get $|a + b|^2 = |c|^2$. This implies $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$. Also,we are given $|a| + |b| = |c|$. Squaring both sides,we get $(|a| + |b|)^2 = |c|^2$. This implies $|a|^2 + |b|^2 + 2|a||b| = |c|^2$. Comparing the two expressions for $|c|^2$,we have $|a|^2 + |b|^2 + 2(a \cdot b) = |a|^2 + |b|^2 + 2|a||b|$. This simplifies to $a \cdot b = |a||b|$. Since $a \cdot b = |a||b| \cos 	heta$,we have $|a||b| \cos 	heta = |a||b|$. Assuming $a$ and $b$ are non-zero vectors,$\cos 	heta = 1$,which means $	heta = 0$.

If $|a| + |b| = |c|$ and $a + b = c$,then the angle between $a$ and $b$ is

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