Vector inner product In △ ABC, | ab-bc | = | AC | = 5, | AB = 2 | BC |, find the area of △ ABC |AB|=2|BC|

Vector inner product In △ ABC, | ab-bc | = | AC | = 5, | AB = 2 | BC |, find the area of △ ABC |AB|=2|BC|

|Ab-bc | = | AC | = 5, that is, | ab-bc | = | AB + BC | = 5, | ab-bc | = | AB + BC | indicates that the diagonal length of a parallelogram with vector AB and vector BC as adjacent sides is equal, and the parallelogram is rectangular, so ∠ B = 90 °. From Pythagorean theorem, | BC | & sup2; + | ab | & sup2; = | AC | & sup2; and because | ab | = 2 | BC |, | AC | =

Arcsinx = 1 / (SiNx) is that right?

You must be influenced by the calculator. Arcsin on the calculator is expressed as sin ^ - 1, not 1 / sin X. This is related to the inverse relation in the group

Unequal relations among SiNx, X and arcsinx
When x approaches 0, is there any unequal relationship among SiNx, X and arcsinx?
Two supplements: 1. Can inequality be equal? 2. Is it also true between (- π, 0) because these three functions are odd

Here we have to make a 0

The relationship between SiNx and arcsinx

These two are inverse functions of each other in the interval [- π / 2, π / 2]