Question

what is the measure of ∠r in △pqr? round to the nearest degree. 26° 42° 45° 64°

Explanation:

Step1: Identify triangle type and sides

$\triangle PQR$ is right - angled at $P$. So, $PQ = 12.7$, $PR = 14.1$? Wait, no, in right - triangle $PQR$ with right angle at $P$, the sides: $PQ = 12.7$, $PR$? Wait, no, the side opposite to $\angle R$ is $PQ = 12.7$, and the hypotenuse? Wait, no, $QR = 14.1$? Wait, no, in right - triangle $PQR$, right - angled at $P$, so $PQ$ and $PR$ are the legs, and $QR$ is the hypotenuse? Wait, no, looking at the diagram, $PQ$ is one leg (length 12.7), $PR$? Wait, no, the right angle is at $P$, so $PQ$ and $PR$ are perpendicular. Wait, the side $QR$ is 14.1? Wait, no, the side adjacent to $\angle R$ is $PR$? Wait, no, let's use trigonometry. In right - triangle $PQR$, $\sin R=\frac{PQ}{QR}$, where $PQ = 12.7$ and $QR = 14.1$? Wait, no, $PQ$ is the side opposite $\angle R$, and $QR$ is the hypotenuse? Wait, no, $PQ = 12.7$, $PR$? Wait, no, the right angle is at $P$, so $PQ$ and $PR$ are legs, and $QR$ is hypotenuse. Wait, $PQ = 12.7$, $QR = 14.1$? Wait, no, maybe $PQ = 12.7$, $QR = 14.1$? Wait, let's check: $\sin R=\frac{PQ}{QR}=\frac{12.7}{14.1}$. Let's calculate that.

Step2: Calculate $\sin R$

$\sin R=\frac{12.7}{14.1}\approx0.9007$

Step3: Find the angle $R$

$R=\sin^{- 1}(0.9007)\approx64^{\circ}$? Wait, no, wait, maybe I mixed up opposite and adjacent. Wait, if the right angle is at $P$, then the sides: $PQ$ is one leg (length 12.7), $PR$ is the other leg, and $QR$ is the hypotenuse (length 14.1)? Wait, no, that can't be, because in a right - triangle, the hypotenuse must be longer than either leg. But 14.1 is longer than 12.7, so that's okay. Wait, but if $PQ = 12.7$ (opposite $\angle R$) and $QR = 14.1$ (hypotenuse), then $\sin R=\frac{PQ}{QR}=\frac{12.7}{14.1}\approx0.9007$, then $R=\sin^{-1}(0.9007)\approx64^{\circ}$? But wait, maybe it's $\cos R=\frac{PR}{QR}$, but we don't know $PR$. Wait, no, maybe I got the sides wrong. Wait, the other leg is $PR$, and $PQ = 12.7$, $QR = 14.1$. Wait, let's check with $\cos R=\frac{PR}{QR}$, but we can find $PR$ using Pythagoras: $PR=\sqrt{QR^{2}-PQ^{2}}=\sqrt{14.1^{2}-12.7^{2}}=\sqrt{(14.1 - 12.7)(14.1 + 12.7)}=\sqrt{1.4\times26.8}=\sqrt{37.52}\approx6.125$. Then $\cos R=\frac{PR}{QR}=\frac{6.125}{14.1}\approx0.4344$, then $R=\cos^{-1}(0.4344)\approx64^{\circ}$? Wait, no, $\cos^{-1}(0.4344)\approx64^{\circ}$? Wait, no, $\cos60^{\circ}=0.5$, $\cos64^{\circ}\approx0.4384$, which is close to 0.4344. So that's correct. Wait, but let's use $\tan R=\frac{PQ}{PR}=\frac{12.7}{6.125}\approx2.073$, and $\tan64^{\circ}\approx2.050$, which is close. So the measure of $\angle R$ is approximately $64^{\circ}$.

Answer:

64°