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MAXImum [283]
3 years ago
13

For positive acute angles AA and B,B, it is known that \tan A=\frac{28}{45}tanA= 45 28 ​ and \cos B=\frac{3}{5}.CosB= 5 3 ​ . Fi

nd the value of \cos(A+B)cos(A+B) in simplest form.
Mathematics
1 answer:
MrRa [10]3 years ago
7 0

Answer:

\cos(A + B) = \frac{23}{265}

Step-by-step explanation:

Given

\tan A=\frac{28}{45}

\cos B=\frac{3}{5}

Required

\cos(A+B)

In trigonometry:

\cos(A+B) = \cos\ A\ cosB - sinA\ sinB

We need to solve for cosA, sinA and sinB

Given that:

\tan A=\frac{28}{45} and the tangent of an angle is a ratio of the opposite side (x) to the adjacent side (y),

So, we have:

x =28 and y = 45

For this angle A, we need t calculate its hypotenuse (z).

Using Pythagoras,

z^2 = x^2 + y^2

z = \sqrt{x^2 + y^2

z = \sqrt{28^2 + 45^2

z = \sqrt{2809

z = 53

From here, we can calculate sin and cos A

\sin A= \frac{opposite}{hypotenuse} and \cos A= \frac{adjacent}{hypotenuse}

\sin A= \frac{x}{z}

\sin A= \frac{28}{53}

\cos A= \frac{y}{z}

\cos A= \frac{45}{53}

To angle B.

Give that

\cos B=\frac{3}{5} and the cosine of an angle is a ratio of the adjacent side (a) to the hypotenuse side (c),

So, we have:

a = 3 and b = 5

For this angle B, we need to calculate its opposite (b).

Using Pythagoras,

c^2 = a^2 + b^2

5^2 = 3^2 + b^2

25 = 9 + b^2

Collect like terms

b^2 = 25 - 9

b^2 = 16

b = \sqrt{16

b = 4

From here, we can calculate sin B

\sin B= \frac{opposite}{hypotenuse}

\sin B = \frac{b}{c}

\sin B = \frac{4}{5}

Recall that:

\cos(A+B) = \cos\ A\ cosB - sinA\ sinB

and

\cos B=\frac{3}{5}      \sin B = \frac{4}{5}       \sin A= \frac{28}{53}      \cos A= \frac{45}{53}

\cos(A + B) = \frac{45}{53} * \frac{3}{5} - \frac{28}{53} * \frac{4}{5}

\cos(A + B) = \frac{135}{265} - \frac{112}{265}

\cos(A + B) = \frac{135-112}{265}

\cos(A + B) = \frac{23}{265}

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