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Question

If 13sinA=5 and A is acute, find the value of 5sinA2cosAtanA.

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Solution

Let PQR be a right angled triangle where Q=900 and R=A as shown in the above figure:

Now it is given that 13sinA=5 that issinA=513

We know that, in a right angled triangle, sinθ is equal to opposite side over hypotenuse that is sinθ=OppositesideHypotenuse, therefore, opposite side PQ=5 and hypotenuse PR=13.

Now, using pythagoras theorem in PQR, we have

PR2=PQ2+QR2132=52+QR2169=25+QR2QR2=16925=144QR=144=12

Therefore, the adjacent side QR=12.

We know that, in a right angled triangle,

cosθ is equal to adjacent side over hypotenuse that is cosθ=AdjacentsideHypotenuse and

tanθ is equal to opposite side over adjacent side that is tanθ=OppositesideAdjacentside

Here, we have opposite side PQ=5, adjacent side QR=12 and the hypotenuse PR=13, therefore, the trignometric ratios of angle A can be determined as follows:

cosA=AdjacentsideHypotenuse=QRPR=1213

tanθ=OppositesideAdjacentside=PQQR=512

Now, we find

5sinA2cosAtanA=(5×513)(2×1213)512=25132413512=113512=113×125=1265

Hence, 5sinA2cosAtanA=1265.

637830_561705_ans.png

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