In a ,triangle ABC, right angled at C, if tan A =1/root 3
find the value of sin A cos B + cos A sin B.
Answers
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." hope it helps Mark as brainlliest pls
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