Question

slove the problem fastly mates no. 6​

Answer 1

A+B=10i^ +5j^=

|A+B|=√(10)²+(5)²=√(125=5√5.

|A|=√(4)²+(-3)²=√25=5

|B|=√(6²+8²=√(100)=10

tan©=|A|/|B|=5/10

©=tan–¹(1/2)

Answer 2

$( 4i - 3j) + (6i + 8j)$

$= 10i + 5j$

magnitude =

$\sqrt {10 {}^{2} + ( 5) {}^{2} } = \sqrt{100 + 25} = \sqrt{125} = 5 \sqrt{5}$

tan@=5/10 =1/2

$\alpha = \tan {}^{ - 1} ( \frac{1}{2} )$