Plz give me solution of 97 and 98 as soon as possible
Attachments:
Answers
Answered by
0
sinA=1/√5
cosA=√(1-sin²A)
cosA=2/√5
and,
sinB=1/√10
cosB=3/√10
now,
cos(A+B)=cosA.cosB-sinA.sinB
cos(A+B)=(2/√5)(3/√10)-(1/√5)(1/√10)
cos(A+B)=(6-1)/√50
cos(A+B)= 5/5√2
cos(A+B)=1/√2=cos45°
A+B=45°
cosA=√(1-sin²A)
cosA=2/√5
and,
sinB=1/√10
cosB=3/√10
now,
cos(A+B)=cosA.cosB-sinA.sinB
cos(A+B)=(2/√5)(3/√10)-(1/√5)(1/√10)
cos(A+B)=(6-1)/√50
cos(A+B)= 5/5√2
cos(A+B)=1/√2=cos45°
A+B=45°
Similar questions