Yasmin
2021-08-14
Answered

In triangle DEF, side E is 4 cm long and side F is 7 cm long. If the angle between sides E and F is 50 degrees, how long is side D?

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Brighton

Answered 2021-08-15
Author has **103** answers

This is a SAS triangle (two sides and an included angle) so we can use the Law of Cosines to find the length of side D:

asked 2022-05-26

If $\alpha +\beta +\gamma =\frac{\pi}{2},$, then prove that

$\frac{(1-\mathrm{tan}\frac{\alpha}{2})(1-\mathrm{tan}\frac{\beta}{2})(1-\mathrm{tan}\frac{\gamma}{2})}{(1+\mathrm{tan}\frac{\alpha}{2})(1+\mathrm{tan}\frac{\beta}{2})(1+\mathrm{tan}\frac{\gamma}{2})}=\frac{\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\gamma -1}{\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma}$

$\frac{(1-\mathrm{tan}\frac{\alpha}{2})(1-\mathrm{tan}\frac{\beta}{2})(1-\mathrm{tan}\frac{\gamma}{2})}{(1+\mathrm{tan}\frac{\alpha}{2})(1+\mathrm{tan}\frac{\beta}{2})(1+\mathrm{tan}\frac{\gamma}{2})}$

$=\frac{1-\mathrm{tan}\frac{\alpha}{2}-\mathrm{tan}\frac{\beta}{2}-\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}+\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\gamma}{2}-\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}}{1+\mathrm{tan}\frac{\alpha}{2}+\mathrm{tan}\frac{\beta}{2}+\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}+\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}}$

I am stuck here

$\frac{(1-\mathrm{tan}\frac{\alpha}{2})(1-\mathrm{tan}\frac{\beta}{2})(1-\mathrm{tan}\frac{\gamma}{2})}{(1+\mathrm{tan}\frac{\alpha}{2})(1+\mathrm{tan}\frac{\beta}{2})(1+\mathrm{tan}\frac{\gamma}{2})}=\frac{\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\gamma -1}{\mathrm{cos}\alpha +\mathrm{cos}\beta +\mathrm{cos}\gamma}$

$\frac{(1-\mathrm{tan}\frac{\alpha}{2})(1-\mathrm{tan}\frac{\beta}{2})(1-\mathrm{tan}\frac{\gamma}{2})}{(1+\mathrm{tan}\frac{\alpha}{2})(1+\mathrm{tan}\frac{\beta}{2})(1+\mathrm{tan}\frac{\gamma}{2})}$

$=\frac{1-\mathrm{tan}\frac{\alpha}{2}-\mathrm{tan}\frac{\beta}{2}-\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}+\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\gamma}{2}-\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}}{1+\mathrm{tan}\frac{\alpha}{2}+\mathrm{tan}\frac{\beta}{2}+\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}+\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\gamma}{2}+\mathrm{tan}\frac{\alpha}{2}\mathrm{tan}\frac{\beta}{2}\mathrm{tan}\frac{\gamma}{2}}$

I am stuck here

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The base of a triangle is 3 ft more than the height. If the area is 14 ft², find the base and the height.

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Solve the equation
$\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}=(\mathrm{sec}x-\mathrm{tan}x{)}^{2}$

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Establishing the identity $\frac{\mathrm{sin}(\alpha +\beta )}{\mathrm{cos}\alpha \mathrm{cos}\beta}=\mathrm{tan}\alpha +\mathrm{tan}\beta $

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Given the cone $z=\sqrt{{x}^{2}+{y}^{2}}$ and the plane z=5+y.

Represent the curve of intersection of the surfaces with a vector function r(t).

Represent the curve of intersection of the surfaces with a vector function r(t).

asked 2022-06-04

By doing some right triangle gymnastics, we can derive things like

$\mathrm{cos}(\mathrm{arctan}x)=\frac{1}{\sqrt{1+{x}^{2}}}$, for $x>0$ $\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

$\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

$\mathrm{tan}(\mathrm{arcsin}x)=\frac{x}{\sqrt{1-{x}^{2}}}$

What about $\mathrm{arctan}\mathrm{cos}(x)$, $\mathrm{arcsin}(\mathrm{tan}x)$, etc?

$\mathrm{cos}(\mathrm{arctan}x)=\frac{1}{\sqrt{1+{x}^{2}}}$, for $x>0$ $\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

$\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

$\mathrm{tan}(\mathrm{arcsin}x)=\frac{x}{\sqrt{1-{x}^{2}}}$

What about $\mathrm{arctan}\mathrm{cos}(x)$, $\mathrm{arcsin}(\mathrm{tan}x)$, etc?