Question

7 practice 7 (from unit 4, lesson 3)
here is triangle pqr.
find the length of each leg.
type your answers in the boxes.
pq = \boxed{} units
qr = \boxed{} units
how did i do?

Explanation:

Step1: Identify the triangle type

Triangle \( PQR \) is a right - triangle with \( \angle Q = 90^{\circ} \), \( \angle P=60^{\circ} \), \( \angle R = 30^{\circ} \) and hypotenuse \( PR = 4 \) units. In a 30 - 60 - 90 triangle, the sides are in the ratio \( 1:\sqrt{3}:2 \), where the side opposite \( 30^{\circ} \) is the shortest side (let's call it \( x \)), the side opposite \( 60^{\circ} \) is \( x\sqrt{3} \), and the hypotenuse is \( 2x \).

Step2: Find the length of \( PQ \)

The side \( PQ \) is opposite the \( 30^{\circ} \) angle (\( \angle R \)). Let the length of \( PQ=x \). Since the hypotenuse \( PR = 4 \) and in a 30 - 60 - 90 triangle, hypotenuse \( = 2x \). So, \( 2x=4 \), then \( x = \frac{4}{2}=2 \). Wait, no. Wait, \( \angle R = 30^{\circ} \), so the side opposite \( \angle R \) is \( PQ \), and the side opposite \( \angle P = 60^{\circ} \) is \( QR \). Wait, let's correct: In right - triangle \( PQR \), right - angled at \( Q \), \( \angle R = 30^{\circ} \), \( \angle P=60^{\circ} \), hypotenuse \( PR = 4 \).

The side opposite \( 30^{\circ} \) (which is \( \angle R \)) is \( PQ \). In a 30 - 60 - 90 triangle, the length of the side opposite \( 30^{\circ} \) is half of the hypotenuse. So, \( PQ=\frac{1}{2}\times PR \). Since \( PR = 4 \), then \( PQ=\frac{1}{2}\times4 = 2 \) units.

Step3: Find the length of \( QR \)

The side \( QR \) is opposite the \( 60^{\circ} \) angle (\( \angle P \)). In a 30 - 60 - 90 triangle, the length of the side opposite \( 60^{\circ} \) is \( \sqrt{3} \) times the length of the side opposite \( 30^{\circ} \). We know that the side opposite \( 30^{\circ} \) ( \( PQ \)) is 2 units. So, \( QR = PQ\times\sqrt{3}=2\times\sqrt{3}=2\sqrt{3} \)? Wait, no. Wait, another way: Using trigonometric ratios.

We can use cosine and sine. For \( PQ \): \( \cos(60^{\circ})=\frac{PQ}{PR} \). Since \( \cos(60^{\circ})=\frac{1}{2} \) and \( PR = 4 \), then \( \frac{1}{2}=\frac{PQ}{4} \), so \( PQ = 4\times\frac{1}{2}=2 \) units.

For \( QR \): \( \sin(60^{\circ})=\frac{QR}{PR} \). Since \( \sin(60^{\circ})=\frac{\sqrt{3}}{2} \) and \( PR = 4 \), then \( QR = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3} \)? Wait, no, wait. Wait, \( \angle P = 60^{\circ} \), so \( \cos(\angle P)=\frac{PQ}{PR} \), \( \sin(\angle P)=\frac{QR}{PR} \). Wait, no: In right - triangle \( PQR \), \( \angle Q = 90^{\circ} \), so:

\( \cos(\angle P)=\frac{PQ}{PR} \), \( \sin(\angle P)=\frac{QR}{PR} \)

\( \cos(60^{\circ})=\frac{PQ}{4} \), \( \cos(60^{\circ}) = 0.5 \), so \( PQ=4\times0.5 = 2 \)

\( \sin(60^{\circ})=\frac{QR}{4} \), \( \sin(60^{\circ})=\frac{\sqrt{3}}{2} \), so \( QR = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3}\approx3.464 \)? Wait, no, wait, maybe I mixed up the angles. Wait, \( \angle R = 30^{\circ} \), so \( \sin(\angle R)=\frac{PQ}{PR} \), \( \cos(\angle R)=\frac{QR}{PR} \)

\( \sin(30^{\circ})=\frac{PQ}{4} \), \( \sin(30^{\circ})=\frac{1}{2} \), so \( PQ = 4\times\frac{1}{2}=2 \)

\( \cos(30^{\circ})=\frac{QR}{4} \), \( \cos(30^{\circ})=\frac{\sqrt{3}}{2} \), so \( QR = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3}\approx3.464 \)? Wait, but let's check again.

Wait, the triangle is right - angled at \( Q \), so:

- \( PQ \) is adjacent to \( \angle P \) and opposite to \( \angle R \)
- \( QR \) is adjacent to \( \angle R \) and opposite to \( \angle P \)
- \( PR \) is the hypotenuse.

Since \( \angle R = 30^{\circ} \), the side opposite \( \angle R \) is \( PQ \). In a 30 - 60 - 90 triangle, the side opposite 30° is half the hypotenuse. So \( PQ=\frac{1}{2}\times PR=\frac{1}{2}\times4 = 2 \) units.

The side opposite \( \angle P = 60^{\circ} \) is \( QR \). In a 30 - 60 - 90 triangle, the side opposite 60° is \( \sqrt{3} \) times the side opposite 30°. Since the side opposite 30° ( \( PQ \)) is 2, then \( QR = 2\times\sqrt{3}=2\sqrt{3}\approx3.464 \)? Wait, no, wait, \( 2\sqrt{3}\approx3.464 \), but let's confirm with trigonometry.

Using \( \cos(\angle P)=\frac{PQ}{PR} \), \( \cos(60^{\circ}) = 0.5=\frac{PQ}{4}\Rightarrow PQ = 2 \)

Using \( \sin(\angle P)=\frac{QR}{PR} \), \( \sin(60^{\circ})=\frac{\sqrt{3}}{2}=\frac{QR}{4}\Rightarrow QR = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3}\approx3.464 \)

Wait, but maybe I made a mistake in the angle - side correspondence. Let's label the triangle:

- Right angle at \( Q \), so \( Q \) is the right - angle vertex.
- \( P \) and \( R \) are the other two vertices.
- \( \angle P = 60^{\circ} \), \( \angle R = 30^{\circ} \)
- Hypotenuse \( PR = 4 \)

So, \( PQ \): adjacent to \( \angle P \), opposite to \( \angle R \)

\( QR \): adjacent to \( \angle R \), opposite to \( \angle P \)

So, for \( PQ \): opposite \( 30^{\circ} \), so length \(=\frac{1}{2}\times\) hypotenuse \(=\frac{1}{2}\times4 = 2 \)

For \( QR \): opposite \( 60^{\circ} \), so length \(=\frac{\sqrt{3}}{2}\times\) hypotenuse \(=\frac{\sqrt{3}}{2}\times4 = 2\sqrt{3}\approx3.464 \)

Wait, but \( 2\sqrt{3}\approx3.464 \), but let's check with Pythagoras theorem.

If \( PQ = 2 \), \( QR = 2\sqrt{3} \), and \( PR = 4 \), then \( PQ^{2}+QR^{2}=2^{2}+(2\sqrt{3})^{2}=4 + 12 = 16 \), and \( PR^{2}=4^{2}=16 \). So it satisfies the Pythagorean theorem (\( a^{2}+b^{2}=c^{2} \), where \( a = PQ \), \( b = QR \), \( c = PR \)). So the values are correct.

Answer:

\( PQ=\boldsymbol{2} \) units, \( QR=\boldsymbol{2\sqrt{3}} \) (or approximately \( 3.46 \)) units. Wait, but the problem says "type your answers in the boxes". Maybe we can write \( PQ = 2 \) and \( QR = 2\sqrt{3}\approx3.46 \), but let's see, in exact form, \( PQ = 2 \), \( QR = 2\sqrt{3} \).