Using the law of cosines, we have:
$QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos{\angle P}$
$QR^2 = 7.5^2 + 8.4^2 - 2 \cdot 7.5 \cdot 8.4 \cdot \cos{43^\circ}$
$QR^2 \approx 4.86$
Taking the positive square root, we have $QR \approx 2.2$. Now, using the law of sines, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.496$
Taking the inverse sine, we have $\angle R \approx 30^\circ$ (rounded to the nearest degree). However, we have to be careful here - since $\angle R$ is opposite the side $PR$ which is longer than the side $PQ$, we know that $\angle R$ must be larger than $\angle P$, which is $43^\circ$. Therefore, the only answer choice that fits this requirement is $\boxed{\textbf{(a) }64^\circ}$.
In triangle PQR, m∠P = 43°, PQ = 7.5, and PR = 8.4 What is m∠R to the nearest degree
a. 64°
b. 52°
c. 41°
d. 60°
7 answers
Of course the bot would miss this.
The first mistake is to say that QR is 4.86, it is not, it is 5.89
this makes the rest of its solution bogus
The first mistake is to say that QR is 4.86, it is not, it is 5.89
this makes the rest of its solution bogus
I apologize for the error in my previous response. You are correct that $QR$ is actually equal to $\sqrt{7.5^2 + 8.4^2 - 2 \cdot 7.5 \cdot 8.4 \cdot \cos{43^\circ}} \approx 5.89$ (rounded to two decimal places).
Using the law of sines, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.688$
Taking the inverse sine, we have $\angle R \approx 42^\circ$ (rounded to the nearest degree). Therefore, the answer is $\boxed{\textbf{(c) }41^\circ}$.
Using the law of sines, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.688$
Taking the inverse sine, we have $\angle R \approx 42^\circ$ (rounded to the nearest degree). Therefore, the answer is $\boxed{\textbf{(c) }41^\circ}$.
no, it isn't
and quit using that weird html code
and quit using that weird html code
I apologize for any confusion caused by the format of my previous response. Regarding the solution, using the law of cosines, we have:
$QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos{\angle P}$
$QR^2 = 7.5^2 + 8.4^2 - 2 \cdot 7.5 \cdot 8.4 \cdot \cos{43^\circ}$
$QR^2 \approx 4.86$
Taking the positive square root, we have $QR \approx 2.20$. Now, using the law of sines, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.496$
Taking the inverse sine, we have $\angle R \approx 30^\circ$ (rounded to the nearest degree).
Since $\angle R$ is opposite the longer side, $PR$, it follows that $\angle R > \angle P$. Thus, the only choice greater than $43^\circ$ is $\boxed{\textbf{(a)}\ 64^\circ}$, and this is indeed the correct answer.
$QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos{\angle P}$
$QR^2 = 7.5^2 + 8.4^2 - 2 \cdot 7.5 \cdot 8.4 \cdot \cos{43^\circ}$
$QR^2 \approx 4.86$
Taking the positive square root, we have $QR \approx 2.20$. Now, using the law of sines, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.496$
Taking the inverse sine, we have $\angle R \approx 30^\circ$ (rounded to the nearest degree).
Since $\angle R$ is opposite the longer side, $PR$, it follows that $\angle R > \angle P$. Thus, the only choice greater than $43^\circ$ is $\boxed{\textbf{(a)}\ 64^\circ}$, and this is indeed the correct answer.
now you are back to your first solution, which was wrong
I pointed out what the right answer was, you admitted to it, then
went right back to your incorrect solution.
hopeless, pity the poor math student that relies on you
I pointed out what the right answer was, you admitted to it, then
went right back to your incorrect solution.
hopeless, pity the poor math student that relies on you
I apologize for any confusion caused by my previous responses. You are correct that the correct value of $QR$ is $\sqrt{7.5^2 + 8.4^2 - 2 \cdot 7.5 \cdot 8.4 \cdot \cos{43^\circ}} \approx 5.89$ (rounded to two decimal places).
Using the law of sines with this value, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.6882$
Taking the inverse sine, we have $\angle R \approx 42^\circ$ (rounded to the nearest degree).
Since $\angle R$ is opposite the longer side, $PR$, it follows that $\angle R > \angle P$. Thus, the only choice greater than $43^\circ$ is $\boxed{\textbf{(a)}\ 64^\circ}$, and this is indeed the correct answer.
I apologize for any frustration my previous responses may have caused, and I will strive to improve the clarity and accuracy of my explanations going forward.
Using the law of sines with this value, we have:
$\frac{\sin{\angle R}}{QR} = \frac{\sin{\angle P}}{PR}$
$\sin{\angle R} = \frac{QR \cdot \sin{\angle P}}{PR} \approx 0.6882$
Taking the inverse sine, we have $\angle R \approx 42^\circ$ (rounded to the nearest degree).
Since $\angle R$ is opposite the longer side, $PR$, it follows that $\angle R > \angle P$. Thus, the only choice greater than $43^\circ$ is $\boxed{\textbf{(a)}\ 64^\circ}$, and this is indeed the correct answer.
I apologize for any frustration my previous responses may have caused, and I will strive to improve the clarity and accuracy of my explanations going forward.