Question

find ( mangle s ).

( mangle s=) 。

Explanation:

Step1: Identify triangle type

Triangle \( RQS \) is right - angled at \( R \), so it is a right - triangle. We know the length of the leg \( SR = 10 \) and the hypotenuse \( SQ=10\sqrt{2}\).

Step2: Use trigonometric ratio or special triangle property

In a right - triangle, we can use the sine function. The sine of an angle \( \theta \) in a right - triangle is defined as \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle S \), the opposite side to \( \angle S \) is \( RQ \) and the adjacent side is \( SR \), and the hypotenuse is \( SQ \). But we can also recognize the triangle by the ratio of sides. In a right - triangle, if the ratio of a leg to the hypotenuse is \( \frac{10}{10\sqrt{2}}=\frac{1}{\sqrt{2}} \), we know that in a \( 45 - 45-90 \) triangle (isosceles right - triangle), the legs are equal and the hypotenuse is \( \text{leg}\times\sqrt{2} \). Since one leg \( SR = 10 \), and if the triangle is isosceles right - triangle, then the other leg \( RQ = 10 \), and the hypotenuse \( SQ = 10\sqrt{2} \), which matches the given lengths. In an isosceles right - triangle, the non - right angles are equal and each is \( 45^{\circ} \). So \( m\angle S = 45^{\circ} \). We can also use the sine formula: \( \sin(\angle S)=\frac{RQ}{SQ} \). Let's assume \( RQ=x \), from Pythagoras theorem \( SR^{2}+RQ^{2}=SQ^{2} \), \( 10^{2}+x^{2}=(10\sqrt{2})^{2} \), \( 100 + x^{2}=200 \), \( x^{2}=100 \), \( x = 10 \). Then \( \sin(\angle S)=\frac{RQ}{SQ}=\frac{10}{10\sqrt{2}}=\frac{1}{\sqrt{2}} \), and we know that \( \sin(45^{\circ})=\frac{1}{\sqrt{2}} \), so \( \angle S = 45^{\circ} \).

Answer:

\( 45 \)