Question

1. which of the following statements is true? figure 3 options: sr = 10√3, pr = 10, pq = 10, ps = 10√3

Explanation:

Step1: Analyze triangle RQS

In right triangle \( RQS \), \( \angle QRS = 30^\circ \), \( SQ = 10 \). In a 30 - 60 - 90 triangle, the sides are in the ratio \( 1:\sqrt{3}:2 \) (opposite 30°, 60°, 90° respectively). The side opposite 30° is \( SQ = 10 \), so the hypotenuse \( RQ \) would be \( 20 \), and the side \( SR \) (opposite 60°) would be \( 10\sqrt{3} \)? Wait, no, wait. Wait, \( \angle Q = 180^\circ - 60^\circ - (90^\circ + 30^\circ) \)? Wait, maybe better to look at triangle \( PRS \) and \( QRS \). Wait, \( \angle P = 60^\circ \), \( \angle PSR = 90^\circ \), so \( \angle PRS = 30^\circ \). Also, \( SQ = 10 \), and since \( RS \perp PQ \), and looking at the angles, triangle \( PQR \) might be isoceles? Wait, no, let's check the options.

Wait, the option \( PR = 10 \)? Wait, no. Wait, let's check each option:

1. \( SR = 10\sqrt{3} \): Let's see, in triangle \( RQS \), \( \angle QRS = 30^\circ \), \( SQ = 10 \). \( \tan(30^\circ)=\frac{SQ}{SR}=\frac{10}{SR} \), so \( SR = \frac{10}{\tan(30^\circ)}=\frac{10}{\frac{1}{\sqrt{3}}}=10\sqrt{3} \)? Wait, no, \( \tan(\theta)=\frac{opposite}{adjacent} \), so \( \tan(30^\circ)=\frac{SQ}{SR} \), so \( SR = \frac{SQ}{\tan(30^\circ)} = 10\sqrt{3} \). Wait, but let's check other options.

2. \( PR = 10 \): In triangle \( PSR \), \( \angle P = 60^\circ \), \( \angle PSR = 90^\circ \), so \( \angle PRS = 30^\circ \). If \( PS = SQ = 10 \) (since \( RS \) is perpendicular and maybe \( PQ \) is bisected? Wait, \( SQ = 10 \), if \( PS = SQ \), then \( PQ = 20 \), but the option \( PQ = 10 \) is wrong. Wait, no, the length \( SQ = 10 \), and if \( PS = SQ \), then \( PS = 10 \), but \( PR \): in triangle \( PSR \), \( \angle P = 60^\circ \), \( \angle PSR = 90^\circ \), so \( \cos(60^\circ)=\frac{PS}{PR} \), if \( PS = 10 \), then \( PR = \frac{PS}{\cos(60^\circ)}=\frac{10}{0.5}=20 \). Wait, maybe I made a mistake.

Wait, the option \( PR = 10 \) is wrong. Wait, the option \( PQ = 10 \): but \( SQ = 10 \) and \( PS \) is some length, so \( PQ = PS + SQ \), so \( PQ > 10 \), so that's wrong.

Option \( PS = 10\sqrt{3} \): In triangle \( PSR \), \( \angle P = 60^\circ \), \( \angle PSR = 90^\circ \), \( \tan(60^\circ)=\frac{SR}{PS} \), if \( SR = 10\sqrt{3} \), then \( \tan(60^\circ)=\sqrt{3}=\frac{10\sqrt{3}}{PS} \), so \( PS = 10 \). So that's wrong.

Wait, the first option \( SR = 10\sqrt{3} \): Let's recalculate. In right triangle \( RQS \), \( \angle QRS = 30^\circ \), \( SQ = 10 \) (opposite \( 30^\circ \) angle). In a 30 - 60 - 90 triangle, the side opposite \( 30^\circ \) is half the hypotenuse, so hypotenuse \( RQ = 20 \), and the side opposite \( 60^\circ \) (which is \( SR \)) is \( 10\sqrt{3} \) (since \( \sin(60^\circ)=\frac{SR}{RQ} \), so \( SR = RQ\sin(60^\circ)=20\times\frac{\sqrt{3}}{2}=10\sqrt{3} \)). So \( SR = 10\sqrt{3} \) is correct? Wait, but let's check the other options again.

Wait, the option \( PR = 10 \): No, because in triangle \( PSR \), \( \angle P = 60^\circ \), \( \angle PSR = 90^\circ \), so \( PS \) is adjacent to \( 60^\circ \), \( PR \) is hypotenuse. If \( PS = 10 \) (since \( SQ = 10 \) and maybe \( PS = SQ \) because of the angles), then \( \cos(60^\circ)=\frac{PS}{PR} \), so \( PR = \frac{PS}{\cos(60^\circ)} = \frac{10}{0.5}=20 \), so \( PR = 20 \), not 10.

Option \( PQ = 10 \): But \( SQ = 10 \) and \( PS \) is some length, so \( PQ = PS + SQ > 10 \), so wrong.

Option \( PS = 10\sqrt{3} \): If \( SR = 10\sqrt{3} \) and \( \tan(60^\circ)=\frac{SR}{PS} \), then \( \sqrt{3}=\frac{10\sqrt{3}}{PS} \), so \( PS = 10 \), so wrong.

So the correct option is \( SR = 10\sqrt{3} \)? Wait, but let's check again. Wait, in triangle \( RQS \), \( \angle QRS = 30^\circ \), \( SQ = 10 \), right angle at \( S \). So \( \tan(30^\circ)=\frac{SQ}{SR} \), so \( SR = \frac{SQ}{\tan(30^\circ)} = 10\sqrt{3} \), which matches the first option.

Wait, but maybe I made a mistake. Wait, the first option is \( SR = 10\sqrt{3} \), let's confirm.

Step2: Verify each option

- Option 1: \( SR = 10\sqrt{3} \): As calculated, using \( \tan(30^\circ)=\frac{SQ}{SR} \), \( SR = 10\sqrt{3} \), correct.
- Option 2: \( PR = 10 \): In triangle \( PSR \), \( \angle P = 60^\circ \), \( \angle PSR = 90^\circ \), so \( PR \) is hypotenuse, \( PS \) is adjacent. If \( PS = 10 \) (assuming \( PS = SQ = 10 \)), then \( PR = \frac{PS}{\cos(60^\circ)} = 20 \), so wrong.
- Option 3: \( PQ = 10 \): \( PQ = PS + SQ \), \( SQ = 10 \), \( PS \) is positive, so \( PQ > 10 \), wrong.
- Option 4: \( PS = 10\sqrt{3} \): From \( \tan(60^\circ)=\frac{SR}{PS} \), \( SR = 10\sqrt{3} \), so \( PS = \frac{SR}{\tan(60^\circ)} = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \), so wrong.

Answer:

\( SR = 10\sqrt{3} \) (the first option)