If cos A=5/13,sin B=4/5.
Find sin(A+B)
tan (A+B)
If A and B are acute angles.
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ARoy
ARoy★ Brainly Teacher ★
cosA=5/13
∴, sinA
=√(1-cos²A)
=√(1-25/169)
=√(169-25)/169
=√(144/169)
=12/13 [neglecting the negative sign since A is an acute angle]
sinB=4/5
∴. cosB
=√(1-sin²B)
=√(1-16/25)
=√(25-16)/25
=√(9/25)
=3/5 [neglecting the negative sign since A is an acute angle]
∴, sin(A+B)
=sinAcosB+cosAsinB
=12/13×3/5+5/13×4/5
=36/65+20/65
=(36+20)/65
=56/65
hope this will help you
Answer:1)56/65
cosA=5/13
∴, sinA
=√(1-cos²A)
=√(1-25/169)
=√(169-25)/169
=√(144/169)
=12/13 [neglecting the negative sign since A is an acute angle]
sinB=4/5
∴. cosB
=√(1-sin²B)
=√(1-16/25)
=√(25-16)/25
=√(9/25)
=3/5 [neglecting the negative sign since A is an acute angle]
∴, sin(A+B)
=sinAcosB+cosAsinB
=12/13×3/5+5/13×4/5
=36/65+20/65
=(36+20)/65
=56/65
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