Question

aliaa and zhang li are tennis - playing robots. aliaa is 715 cm from zhang li and 2,200 cm from the far corner. zhang li measures an 91° angle between aliaa and the far corner, as shown. from aliaa’s viewpoint, what is the angle between zhang li and the far corner? do not round during your calculations. round your final answer to the nearest degree.

Explanation:

Step1: Identify the Law to Use

We have a triangle with sides: \( a = 715 \) cm (Aliaa to Zhang Li), \( b = 2200 \) cm (Aliaa to far corner), and angle \( A = 91^\circ \) (Zhang Li's angle between Aliaa and far corner). We need to find angle \( B \) (Aliaa's angle between Zhang Li and far corner). Use the Law of Sines: \(\frac{\sin A}{a}=\frac{\sin B}{b}\).

Step2: Rearrange the Formula

From \(\frac{\sin A}{a}=\frac{\sin B}{b}\), solve for \(\sin B\): \(\sin B=\frac{b\sin A}{a}\).

Step3: Substitute Values

Substitute \( a = 715 \), \( b = 2200 \), \( A = 91^\circ \):
\(\sin B=\frac{2200\times\sin(91^\circ)}{715}\).
Calculate \(\sin(91^\circ)\approx0.999847695\), so:
\(\sin B=\frac{2200\times0.999847695}{715}=\frac{2199.664929}{715}\approx3.076454446\). Wait, that can't be. Wait, no—wait, I mixed up the sides. Wait, in Law of Sines, \( \frac{\sin A}{a}=\frac{\sin B}{b} \), where \( A \) is opposite side \( a \), \( B \) opposite side \( b \). Wait, actually, let's correct: Let’s define the triangle: Let Aliaa be \( A \), Zhang Li be \( Z \), far corner be \( F \). So sides: \( AZ = 715 \), \( AF = 2200 \), angle at \( Z \) (Zhang Li) is \( \angle AZF = 91^\circ \). We need angle at \( A \) (Aliaa): \( \angle ZAF \). So sides: \( AZ = 715 \) (side opposite \( \angle F \)), \( AF = 2200 \) (side opposite \( \angle Z \)), \( ZF \) is unknown. Wait, no—Law of Sines: \( \frac{\sin \angle Z}{ZF}=\frac{\sin \angle A}{ZF} \)? No, wait, correct labels: Let’s denote:
- Vertex \( A \): Aliaa
- Vertex \( Z \): Zhang Li
- Vertex \( F \): Far corner

So:
- \( AZ = 715 \) (side \( f \), opposite \( \angle F \))
- \( AF = 2200 \) (side \( z \), opposite \( \angle Z \))
- \( \angle Z = 91^\circ \) (angle at \( Z \), between \( A \) and \( F \))
- We need \( \angle A \) (angle at \( A \), between \( Z \) and \( F \))

Law of Sines: \(\frac{\sin \angle A}{ZF}=\frac{\sin \angle Z}{AF}=\frac{\sin \angle F}{AZ}\). Wait, no, better: \(\frac{\sin \angle A}{ZF}=\frac{\sin \angle Z}{AF}\), but we don't know \( ZF \). Wait, no, I made a mistake earlier. Let's re-express: The sides are \( AZ = 715 \), \( AF = 2200 \), and angle at \( Z \) is \( 91^\circ \). So in triangle \( AZF \), we have side \( AZ = 715 \), side \( AF = 2200 \), angle at \( Z \) is \( 91^\circ \). Wait, no—angle at \( Z \) is between \( A \) and \( F \), so the sides adjacent to angle \( Z \) are \( ZA = 715 \) and \( ZF \) (unknown), and the side opposite angle \( Z \) is \( AF = 2200 \). Wait, that's the Law of Sines: \(\frac{\sin \angle A}{ZF}=\frac{\sin \angle Z}{AF}=\frac{\sin \angle F}{AZ}\). But we need angle at \( A \), so let's use \(\frac{\sin \angle A}{ZF}=\frac{\sin \angle Z}{AF}\), but we don't know \( ZF \). Wait, no, I think I mixed up the sides. Wait, the correct Law of Sines is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \( a \) is opposite angle \( A \), etc. So let's assign:
- Let \( a = AF = 2200 \) (opposite angle \( Z \) (angle at \( Z \), which is \( 91^\circ \)))
- Let \( z = AZ = 715 \) (opposite angle \( F \))
- Let \( f = ZF \) (opposite angle \( A \))

Wait, no, angle at \( Z \) is \( \angle AZF = 91^\circ \), so the side opposite angle \( Z \) is \( AF = 2200 \). The side opposite angle \( A \) is \( ZF \), and the side opposite angle \( F \) is \( AZ = 715 \). So Law of Sines: \(\frac{AF}{\sin \angle Z}=\frac{AZ}{\sin \angle F}=\frac{ZF}{\sin \angle A}\). We need angle \( A \) (angle at \( A \), between \( Z \) and \( F \)). Wait, maybe I should use Law of Sines as \(\frac{\sin \angle A}{ZF}=\frac{\sin \angle Z}{AF}\), but we don't know \( ZF \). Wait, no, perhaps I made a mistake in the initial side assignment. Let's start over.

The triangle has:
- Aliaa (A) to Zhang Li (Z): 715 cm (side AZ)
- Aliaa (A) to far corner (F): 2200 cm (side AF)
- Zhang Li (Z) to far corner (F): unknown (side ZF)
- Angle at Z (between A and F): 91° (angle AZF)
- We need angle at A (between Z and F): angle ZAF

So in triangle AZF, we have:
- Side AZ = 715 (adjacent to angle A and angle Z)
- Side AF = 2200 (adjacent to angle A and angle F)
- Angle at Z: 91°

Using Law of Sines: \(\frac{\sin \angle A}{ZF}=\frac{\sin \angle Z}{AF}=\frac{\sin \angle F}{AZ}\)

But we can also write \(\frac{\sin \angle A}{ZF}=\frac{\sin 91^\circ}{2200}\) and \(\frac{\sin \angle F}{715}=\frac{\sin 91^\circ}{2200}\), but we need angle A. Wait, maybe I should use Law of Sines as \(\frac{AZ}{\sin \angle F}=\frac{AF}{\sin \angle Z}\), so \(\sin \angle F=\frac{AZ \times \sin \angle Z}{AF}\). Then, since the sum of angles in a triangle is 180°, angle A = 180° - angle Z - angle F.

Yes, that makes sense. Let's do that.

Step3: Calculate Angle F

\(\sin \angle F=\frac{AZ \times \sin \angle Z}{AF}=\frac{715 \times \sin 91^\circ}{2200}\)

Calculate \(\sin 91^\circ \approx 0.999847695\)

So \(\sin \angle F=\frac{715 \times 0.999847695}{2200}=\frac{714.891102}{2200}\approx0.324950501\)

Step4: Find Angle F

\(\angle F=\arcsin(0.324950501)\approx18.96^\circ\) (using calculator, since \(\sin^{-1}(0.32495)\approx18.96^\circ\))

Step5: Calculate Angle A

Sum of angles in triangle: \( \angle A + \angle Z + \angle F = 180^\circ \)

So \( \angle A = 180^\circ - 91^\circ - 18.96^\circ = 70.04^\circ \), which rounds to 70°. Wait, but let's check the Law of Sines again. Wait, maybe I mixed up the sides. Wait, AZ is 715, AF is 2200, angle at Z is 91°. So side opposite angle Z is AF (2200), side opposite angle A is ZF, side opposite angle F is AZ (715). So Law of Sines: \(\frac{AF}{\sin \angle Z}=\frac{AZ}{\sin \angle F}\), so \(\sin \angle F=\frac{AZ \times \sin \angle Z}{AF}\), which is what I did. Then angle A is 180 - 91 - angle F. Let's recalculate \(\sin \angle F\):

715 * sin(91°) = 715 * 0.999847695 ≈ 715 * 0.9998477 ≈ 714.8911

714.8911 / 2200 ≈ 0.32495

arcsin(0.32495) ≈ 18.96°, so angle A = 180 - 91 - 18.96 = 70.04°, which rounds to 70°. Wait, but let's check with Law of Sines for angle A. Wait, maybe I made a mistake in the side labels. Let's use Law of Sines correctly:

In triangle AZF:
- Side AZ = 715 (length from A to Z)
- Side AF = 2200 (length from A to F)
- Angle at Z (∠AZF) = 91°
- We need angle at A (∠ZAF)

Law of Sines: \(\frac{\sin \angle ZAF}{ZF}=\frac{\sin \angle AZF}{AF}\)

But we can also use \(\frac{\sin \angle ZAF}{\sin \angle AZF}=\frac{ZF}{AF}\), but we don't know ZF. Alternatively, use Law of Sines as \(\frac{AZ}{\sin \angle AFZ}=\frac{AF}{\sin \angle AZF}\), so \(\sin \angle AFZ=\frac{AZ \times \sin \angle AZF}{AF}\), which is what I did. Then angle at A is 180 - 91 - angle AFZ. So that's correct. So angle at A is approximately 70 degrees.

Answer:

70