6. in the triangle, the ratio of the measures of the sides is 10:11:15. what is the measure of each angle?

Question

Accepted Answer

# Explanation:
## Step1: Let the sides be $10x$, $11x$, $15x$
Let the sides of the triangle be $a = 10x$, $b=11x$, $c = 15x$

## Step2: Use the Law of Cosines $c^{2}=a^{2}+b^{2}-2ab\cos C$
For angle $C$ (opposite side $c$):
$$
\begin{align*}
(15x)^{2}&=(10x)^{2}+(11x)^{2}-2\times(10x)\times(11x)\cos C\
225x^{2}&=100x^{2}+121x^{2}-220x^{2}\cos C\
225x^{2}&=221x^{2}-220x^{2}\cos C\
225x^{2}-221x^{2}&=- 220x^{2}\cos C\
4x^{2}&=-220x^{2}\cos C\
\cos C&=\frac{4x^{2}}{- 220x^{2}}=-\frac{1}{55}\approx - 0.0182\
C&\approx91^{\circ}
\end{align*}
$$

## Step3: Use the Law of Cosines $a^{2}=b^{2}+c^{2}-2bc\cos A$
For angle $A$ (opposite side $a$):
$$
\begin{align*}
(10x)^{2}&=(11x)^{2}+(15x)^{2}-2\times(11x)\times(15x)\cos A\
100x^{2}&=121x^{2}+225x^{2}-330x^{2}\cos A\
100x^{2}&=346x^{2}-330x^{2}\cos A\
100x^{2}-346x^{2}&=-330x^{2}\cos A\
- 246x^{2}&=-330x^{2}\cos A\
\cos A&=\frac{-246x^{2}}{-330x^{2}}=\frac{41}{55}\approx0.7455\
A&\approx42^{\circ}
\end{align*}
$$

## Step4: Use the angle - sum property of a triangle $A + B + C=180^{\circ}$
Since $A + B + C = 180^{\circ}$, then $B=180^{\circ}-(A + C)$
$B = 180^{\circ}-(42^{\circ}+91^{\circ}) = 47^{\circ}$

# Answer:
The measures of the angles are approximately $42^{\circ}$, $47^{\circ}$, and $91^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Let the sides be \(10x\), \(11x\), \(15x\)

Step2: Use the Law of Cosines \(c^{2}=a^{2}+b^{2}-2ab\cos C\)

Step3: Use the Law of Cosines \(a^{2}=b^{2}+c^{2}-2bc\cos A\)

Step4: Use the angle - sum property of a triangle \(A + B + C=180^{\circ}\)

Answer: