[Math] Calculate all Angles of the Triangle (all 3 sides given)
trigonometry
so id like to calculate the 3 given Angles as efficient as possible with trigometrics. Maybe with the law of Cosine. So if someone knows the formula for this id be happy to know.
Thanks for help
Best Answer
The largest angle is opposite the longest side. To find the largest angle, use the Law of Cosines
$$c^2 = a^2 + b^2 - 2ab\cos\gamma$$
with $a = 7~\text{cm}$, $b = 5~\text{cm}$, and $c = 10~\text{cm}$.
Once you have solved for $\gamma$, use the Law of Sines
$$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}$$
to determine $\alpha$. Keep in mind that there is only one possible value for $\alpha$ since $\alpha < \gamma$.
Once you have determined both $\alpha$ and $\gamma$, you can obtain $\beta$ by using the Angle Sum Theorem for Triangles, $\alpha + \beta + \gamma = 180^\circ$.
The lengths don't determine the angles: for example with all sides equal you have a rhombus, where the angle at any vertex could be anything in the interval $(0, \pi)$.
Let $Y$ be the rest of the leg on which $A$ lies. Then
$$\begin{align}
\frac{C^2}{Y^2} = \frac{B^2}{B^2 + (A+Y)^2}
&\iff B^2Y^2 = C^2(B^2 + A^2 + 2AY + Y^2)\\
&\iff (B^2 - C^2)Y^2 - 2AC^2Y - C^2(A^2 + B^2) = 0.\end{align}$$
Using the quadratic formula,
$$\begin{align}
Y & = \dfrac{2AC^2 \pm \sqrt{4A^2C^4 + 4(B^2-C^2)C^2(A^2+B^2)}}{2(B^2-C^2)}
\\ & = \dfrac{AC^2 \pm \sqrt{A^2C^4 + A^2B^2C^2 + B^4C^2 - A^2C^4 - B^2C^4}}{B^2-C^2}
\\ &= \dfrac{AC^2 \pm C \sqrt{A^2B^2 + B^4 - B^2C^2}}{B^2-C^2}
\\ &= \dfrac{AC^2 \pm BC \sqrt{A^2 + B^2 - C^2}}{B^2-C^2}.
\end{align}$$
We know $B>C$ since $C$'s triangle is similar to the overall triangle and embedded in it, so the radical is always real. Also $BC \sqrt{A^2 + B^2 - C^2} > ABC > AC^2$, and $Y$ is positive, so the plus case gives us the only solution:
$$Y = \dfrac{AC^2 + BC \sqrt{A^2 + B^2 - C^2}}{B^2-C^2}.$$
Then $\dfrac{Y}{A+Y} = \dfrac{X}{B}$, so $X = \dfrac{BY}{A+Y}$.
Best Answer
The largest angle is opposite the longest side. To find the largest angle, use the Law of Cosines $$c^2 = a^2 + b^2 - 2ab\cos\gamma$$ with $a = 7~\text{cm}$, $b = 5~\text{cm}$, and $c = 10~\text{cm}$.
Once you have solved for $\gamma$, use the Law of Sines $$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}$$ to determine $\alpha$. Keep in mind that there is only one possible value for $\alpha$ since $\alpha < \gamma$.
Once you have determined both $\alpha$ and $\gamma$, you can obtain $\beta$ by using the Angle Sum Theorem for Triangles, $\alpha + \beta + \gamma = 180^\circ$.