Question

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 ​ . Find the value of \cos(A+B)cos(A+B) in simplest form.

Answer 1

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

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

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

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

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

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

Answer 2

The value of $\cos (A+B)$ is $-\frac{96}{265}$.

Trigonometrically speaking, the sine, cosine and tangent of an angle are described by the following formulas:

$\sin \theta = \frac{y}{\sqrt{x^{2}+y^{2}}}$ (1)

$\cos \theta = \frac{x}{\sqrt{x^{2}+y^{2}}}$ (2)

$\tan \theta = \frac{y}{x}$ (3)

Where:

• $\theta$ - Angle, in sexagesimal degrees.
• $x$ - Adjacent leg.
• $y$ - Opposite leg.

And we have the following trigonometric identity for the cosine of the sum of angles:

$\cos (A+B) = \cos A\cdot \cos B - \sin A \cdot \sin B$ (4)

By (1) and (2), we rewrite (4):

$\cos (A+B) = \frac{x_{A}\cdot x_{B}}{\sqrt{x_{A}^{2}+y_{A}^{2}}\cdot \sqrt{x_{B}^{2}+y_{B}^{2}}} - \frac{y_{A}\cdot y_{B}}{\sqrt{x_{A}^{2}+y_{A}^{2}}\cdot \sqrt{x_{B}^{2}+y_{B}^{2}}}$

If we know that $x_{A} = 28$, $y_{A} = 45$, $x_{B} = 3$ and $y_{B} = 4$, then the cosine for the sum of angles is:

$\cos (A+B) = \frac{28\cdot 3}{\sqrt{28^{2}+45^{2}}\cdot \sqrt{3^{2}+4^{2}}}-\frac{45\cdot 4}{\sqrt{28^{2}+45^{2}}\cdot \sqrt{3^{2}+4^{2}}}$

$\cos (A+B) = -\frac{96}{265}$

The value of $\cos (A+B)$ is $-\frac{96}{265}$.

We kindly invite to check this question on trigonometric identities: https://brainly.com/question/24836845