Question

if sin(A+B)=1 and cos B=1/3,then find √A and √B(0°<A+≤90°)​

Answer 1

Correct Question:—

If sin(A+B)=1 and cos B=1/2,then find √A and √B. (0°≤ A and B ≤90°)

$\Huge{\color{red}{\fbox{\textsf{\textbf{♣Solution♣:-}}}}}$

Given,

• sin(A+B)=1
• $\: \: \: \: \: \: \: \: \implies sin(A+B)= \: sin \: {90}^{0}$

$\: \: \: \: \: \implies \: A+B= {90}^{0} \: \: \: - - - - - (1)$

• cosB = 1/2

$\: \: \: \: \: \implies \: cosB = cos {60}^{0}$

$\: \: \: \: \: \: \implies \: B = {60}^{0}$

$\bold { \mathtt{Now, \: \: putting \: \: the \: value \: of \: B \: in \: (1)}}$

$\: \: \implies \: A \: = {30}^{0}$

$\large \bold { \mathtt{So,}}$

• A → 30⁰

$\implies \: \sqrt{A} = \sqrt{{30}^{0} }$

$\implies \: \sqrt{A} = 5.48 \: \: (approx)$

• B → 60⁰

$\implies \: \sqrt{B} = \sqrt{ {60}^{0} }$

$\implies \: \sqrt{B} = 2\sqrt{15}$