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Question

In triangle ABC, right-angled at B, if sin A =1213, then find the value of

(sinA cosA + sinC cosC ).



A

120169 

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B

60169 

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C

1213 

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D
1312 
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Solution

The correct option is B

120169 



SinA=BCAC =1213

Let BC = 12k and AC = 13k, where k is a positive number.

Using Pythagoras Theorem in Triangle ABC, we get,

AC2=BC2+AB2 

(13k)2=(12k)2+(AB)2 

AB2=169k2144k2=25k2

⇒AB = 5k

So, now we can get the required trigonometric ratios:

 cos A = ABAC=5k13k=513

 sin C  = ABAC=5k13k=513

 cos C = BCAC=12k13k=1213

 ⇒ sinA cosA + sin C cos C =   1213×513+513×1213

 =   60169+60169=120169


Mathematics

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