Answer:
[tex]\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})[/tex]
Step-by-step explanation:
Let the hypotenuse of the smaller triangle be h units.
Then; from the Pythagoras Theorem.
[tex]h^2=4^2+2^2[/tex]
[tex]h^2=16+4[/tex]
[tex]h^2=20[/tex]
[tex]h=\sqrt{20}[/tex]
[tex]h=2\sqrt{5}[/tex]
From the smaller triangle;
[tex]\cos (\alpha)=\frac{4}{2\sqrt{5} }=\frac{2}{\sqrt{5} }[/tex] and [tex]\sin(\alpha)=\frac{2}{2\sqrt{5} }=\frac{1}{\sqrt{5} }[/tex]
From the second triangle, let the other other shorter leg of the second triangle be s units.
Then;
[tex]s^2+4^2=6^2[/tex]
[tex]s^2+16=36[/tex]
[tex]s^2=36-16[/tex]
[tex]s^2=20[/tex]
[tex]s=\sqrt{20}[/tex]
[tex]s=2\sqrt{5}[/tex]
[tex]\cos(\beta)=\frac{2\sqrt{5} }{6}=\frac{\sqrt{5} }{3}[/tex]
and
[tex]\sin(\beta)=\frac{4}{6}=\frac{2}{3}[/tex]
We now use the double angle property;
[tex]\cos(\alpha +\beta)=\cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta)[/tex]
we plug in the values to obtain;
[tex]\cos(\alpha +\beta)=\frac{2}{\sqrt{5} }\times \frac{\sqrt{5} }{3}-\frac{1}{\sqrt{5} }\times \frac{2}{3}[/tex]
[tex]\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})[/tex]
Answer:
[tex]cos(\alpha + \beta)=cos(68.4\°) \approx 0.37[/tex]
Step-by-step explanation:
We know both legs, we can use the tangent trigonometric reason to find the angle.
[tex]tan\alpha =\frac{2}{4}\\ tan \alpha=\frac{1}{2}\\ \alpha=tan^{-1}(\frac{1}{2} )\\ \alpha \approx 26.6\°[/tex]
We know the hypothenuse and the opposite leg. We can use the sin trigonometric reason to find the angle
[tex]sin\beta =\frac{4}{6}\\ sin\beta=\frac{2}{3}\\ \beta=sin^{-1} (\frac{2}{3} )\\\beta= 41.8\°[/tex]
So, the sum of them is
[tex]\alpha + \beta = 26.6+41.8= 68.4\°[/tex]
Then,
[tex]cos(\alpha + \beta)=cos(68.4\°) \approx 0.37[/tex]
Therefore,
[tex]cos(\alpha + \beta)=cos(68.4\°) \approx 0.37[/tex]
