Problem 30: We are given a right triangle $PQR$ with $\angle PQR = 90^\circ$, $|QR| = 2$ cm, and $\angle PRQ = 60^\circ$. We need to find the length of the hypotenuse $|PR|$. Problem 31: The interior angles of a triangle are in the ratio $2:5:8$. We need to find the difference between the smallest and largest angles.

1. Problem Description

Problem 30: We are given a right triangle $PQR$ with $\angle PQR = 90^\circ$, $|QR| = 2$ cm, and $\angle PRQ = 60^\circ$. We need to find the length of the hypotenuse $|PR|$.

Problem 31: The interior angles of a triangle are in the ratio $2:5:8$. We need to find the difference between the smallest and largest angles.

2. Solution Steps

Problem 30:

In the right triangle $PQR$, we have $\angle PRQ = 60^\circ$. We can use trigonometric ratios to relate the sides and angles. Specifically, we can use the cosine function:

$$\cos(\angle PRQ) = \frac{|QR|}{|PR|}$$

We are given $|QR| = 2$ cm and $\angle PRQ = 60^\circ$. Therefore,

$$\cos(60^\circ) = \frac{2}{|PR|}$$

We know that $\cos(60^\circ) = \frac{1}{2}$. So,

$$\frac{1}{2} = \frac{2}{|PR|}$$

$$|PR| = 2 \times 2 = 4 \text{ cm}$$

Problem 31:

Let the angles be $2x$, $5x$, and $8x$. The sum of the interior angles of a triangle is $180^\circ$. So,

$$2x + 5x + 8x = 180^\circ$$

$$15x = 180^\circ$$

$$x = \frac{180^\circ}{15} = 12^\circ$$

The angles are $2x = 2(12^\circ) = 24^\circ$, $5x = 5(12^\circ) = 60^\circ$, and $8x = 8(12^\circ) = 96^\circ$.

The smallest angle is $24^\circ$, and the largest angle is $96^\circ$.

The difference between the smallest and largest angles is:

$$96^\circ - 24^\circ = 72^\circ$$

3. Final Answer

Problem 30: B. 4 cm

Problem 31: D. 72°