Question

in triangle abc right angled at c if sin a = 1/2 then find value of sin a cos b + cos a sin b​

Answer 1

Answer:

1/2 only

because when above is half

so it's also half also

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

The value of\sin A \cos B + \cos A \sin B = 1sinAcosB+cosAsinB=1

Given:

\tan A = \frac { 1 } { \sqrt { 3 } }tanA=

3

1

To find:

Value of \sin A \cos B + \cos A \sin BValueofsinAcosB+cosAsinB

Solution:

In the given right angled triangle

\tan A = \frac { 1 } { \sqrt { 3 } }tanA=

3

1

Since \tan 30 ^ { \circ} = \frac { 1 } { \sqrt { 3 } }tan30

∘

=

3

1

A = 30

Given C = 90

So, “A” + “B” + “C” = “180”

or, “B = 60”

Hence, \tan B = \tan 60 ^ { \circ} = \sqrt { 3 }tanB=tan60

∘

=

3

B = 60

Now, \sin A \cos B + \cos A \sin B = \sin ( A + B ) = \sin ( 30 ^ { \circ} + 60 ^ { \circ} ) = \sin 90 ^ { \circ} = 1sinAcosB+cosAsinB=sin(A+B)=sin(30

∘

+60

∘

)=sin90

∘

=1

Hence, the value will be 1."