Question

Please somebody answer from question 11...I am too tired to do it on my own...please somebody​

Answer 1

Answer:

$\sin(a + b) = 1 \\( a + b) = \sin {}^{ - 1} (1) = 90 \\ (a + b) = 90.......(1) \\ \cos(a - b) = 1 \\ (a - b) = \cos {}^{ - 1} (1) = 180 \\ (a - b) = 180.......(2) \\ from \: solving \: of\: (1) \: and \: (2) \: we \: will \: get\\ 2a = 270 \: \: = > a = 135 \\ thus \: \: b = - 45 \\ thus \: its \: follows \: the \: given \: condition \: as \\0 < ( a + b ) \leqslant 90 \\ and \: a > b \\ so \: the \: answer \: is \\ a \: = 135 \\ b = - 45$

similar method for question no. 11

for question no 12 put the value as sin(45 - 30) and cos(45 - 30) instead of 15 where a = 45 and b = 30 and follow the method as given in question. now put the value of all quantities and try to simplify you will get your answer

Answer 2

Answer:

A=30

B=15

Step-by-step explanation:

sin(A+2B)=√3/2=sin=sin60

So,A+2B=60.....................................(i)

cos(A+4B)=0=cos90

So,A+4B=90...............................................................(ii)

(ii)-(i)

2B=30

B=15

Using (i)

A+2*15=60

A=60-30

A=30