Question

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Answer 1

Answer:

SWER

Given=

>5cosA + 12sinA + 12

=>13{\dfrac{5}{13}CosA

13

5

CosA + \dfrac{12}{13}SinA

13

12

SinA } + 12

Let, Cos \theta=\dfrac{5}{13}θ=

13

5

then Sin\theta=\dfrac{12}{13}θ=

13

12

So we get =>

=>13(CosA × Cos\theta + SinA \times Sin\thetaCosθ+SinA×Sinθ ) +12

=>13{Cos(A - \theta)(A−θ) }

Here the minimum value of {Cos(A -\theta )(A−θ) } is -1.

Hence the minimum value of 5cosA + 12sinA + 12 will be -13 + 12=-1.

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Answer 2

Answer:

Step-by-step explanation:

Given= >5cosA + 12sinA + 12

=>13{\dfrac{5}{13}CosA

13

5

CosA + \dfrac{12}{13}SinA

13

12

sinA } + 12

Let, Cos \theta=\dfrac{5}{13}θ=

13

5

hen Sin\theta=\dfrac{12}{13}θ=

13

12

So we get =>

=>13(CosA × Cos\theta + SinA \times Sin\thetaCosθ+SinA×Sinθ ) +12

=>13{Cos(A - \theta)(A−θ) }

Here the minimum value of {Cos(A -\theta )(A−θ) } is -1.

Hence the minimum value of 5cosA + 12sinA + 12 will be -13 + 12=-1.