Question

current attempt in progress
solve the following triangle (if possible).
α = 35°, β = 40°, a = 10
round your answers to one decimal place.
enter na in each answer area if a unique triangle does not exist.
γ =
b ≈
c ≈

Explanation:

Step1: Find angle $\gamma$

The sum of angles in a triangle is $180^{\circ}$. So, $\gamma=180^{\circ}-\alpha - \beta$.
Substituting $\alpha = 35^{\circ}$ and $\beta=40^{\circ}$, we get $\gamma=180^{\circ}-35^{\circ}-40^{\circ}=105^{\circ}$.

Step2: Use the Law of Sines to find side $b$

The Law of Sines states that $\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}$.
We know $a = 10$, $\alpha=35^{\circ}$, and $\beta = 40^{\circ}$.
So, $b=\frac{a\sin\beta}{\sin\alpha}$.
Substituting the values: $b=\frac{10\times\sin40^{\circ}}{\sin35^{\circ}}$.
Since $\sin40^{\circ}\approx0.643$ and $\sin35^{\circ}\approx0.574$, $b=\frac{10\times0.643}{0.574}\approx11.2$.

Step3: Use the Law of Sines to find side $c$

Using the Law of Sines $\frac{a}{\sin\alpha}=\frac{c}{\sin\gamma}$.
We know $a = 10$, $\alpha=35^{\circ}$, and $\gamma = 105^{\circ}$.
So, $c=\frac{a\sin\gamma}{\sin\alpha}$.
Since $\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966$ and $\sin35^{\circ}\approx0.574$.
$c=\frac{10\times0.966}{0.574}\approx16.8$.

Answer:

$\gamma = 105^{\circ}$
$b\approx11.2$
$c\approx16.8$