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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 A

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


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