Question

If cos A=5/13,sin B=4/5.
Find sin(A+B)
tan (A+B)
If A and B are acute angles.​

Answer 1

Step-by-step explanation:

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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 2

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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