Question

consider right triangle △jkl below.
which expressions represent the length of side \\(\overline{kl}\\)?
choose 2 answers:
a \\(\frac{3}{\sin(60^{\circ})}\\)
b \\(\frac{5.2}{\sin(60^{\circ})}\\)
c \\(\frac{3}{\sin(90^{\circ}-60^{\circ})}\\)
d \\(\frac{5.2}{\sin(90^{\circ}-60^{\circ})}\\)

Explanation:

Step1: Recall Sine Law in Triangles

In a right triangle (or any triangle), the Sine Law states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, where $a,b,c$ are sides opposite angles $A,B,C$ respectively. In right triangle $\triangle JKL$, $\angle J = 90^\circ$, $\angle L=60^\circ$, so $\angle K=90^\circ - 60^\circ = 30^\circ$. Side $JL = 3$ (opposite $\angle K$), side $JK = 5.2$ (opposite $\angle L$), and side $KL$ is the hypotenuse (opposite $\angle J = 90^\circ$).

Step2: Apply Sine Law for $\angle L$ and $\angle J$

For angle $L = 60^\circ$ and its opposite side $JK = 5.2$, and hypotenuse $KL$ (opposite right angle, but also we can use angle $L$ and side $JL$). Wait, let's correct: $\angle L = 60^\circ$, opposite side is $JK = 5.2$; $\angle K=30^\circ$, opposite side is $JL = 3$; hypotenuse $KL$. Using Sine Law: $\frac{KL}{\sin 90^\circ}=\frac{JL}{\sin 60^\circ}=\frac{JK}{\sin 30^\circ}$. Since $\sin 90^\circ = 1$, $KL=\frac{JL}{\sin 60^\circ}=\frac{3}{\sin 60^\circ}$ (which is option A). Also, $\sin(90^\circ - 60^\circ)=\sin 30^\circ$, and $\frac{KL}{\sin 90^\circ}=\frac{JK}{\sin(90^\circ - 60^\circ)}$, so $KL=\frac{JK}{\sin(90^\circ - 60^\circ)}=\frac{5.2}{\sin(90^\circ - 60^\circ)}$? Wait no, wait $\angle K = 90 - 60 = 30^\circ$, so $\sin(90 - 60)=\sin 30^\circ$, and side opposite $\angle K$ is $JL = 3$, no. Wait, maybe another approach: $\sin(90^\circ - 60^\circ)=\cos 60^\circ$, but also $\sin(90^\circ - \theta)=\cos \theta$, and $\sin \theta = \cos(90^\circ - \theta)$. Wait, let's re - express:

We know that $\sin(90^\circ - 60^\circ)=\sin 30^\circ$, and from Sine Law, $\frac{JL}{\sin(90^\circ - 60^\circ)}=\frac{KL}{\sin 90^\circ}$, so $KL=\frac{JL}{\sin(90^\circ - 60^\circ)}$? Wait no, $JL$ is opposite $\angle K = 30^\circ$, so $\frac{JL}{\sin \angle K}=\frac{KL}{\sin \angle J}$. $\angle J = 90^\circ$, $\sin \angle J = 1$, $\angle K = 30^\circ=\ 90 - 60$, so $\frac{3}{\sin(90^\circ - 60^\circ)}=\frac{KL}{1}$, so $KL=\frac{3}{\sin(90^\circ - 60^\circ)}$? Wait no, earlier mistake. Wait, $\angle K = 30^\circ$, opposite side $JL = 3$; $\angle L = 60^\circ$, opposite side $JK = 5.2$; hypotenuse $KL$. So $\frac{KL}{\sin 90^\circ}=\frac{JL}{\sin \angle K}=\frac{JK}{\sin \angle L}$. So $\frac{KL}{1}=\frac{3}{\sin 30^\circ}=\frac{5.2}{\sin 60^\circ}$. Since $\sin 30^\circ=\sin(90 - 60)$, so $\frac{3}{\sin(90 - 60)} = KL$ (option C? Wait no, option C is $\frac{3}{\sin(90 - 60)}$, and option D is $\frac{5.2}{\sin(90 - 60)}$. Wait, let's recalculate:

Wait $\angle K = 90 - 60 = 30^\circ$, so $\sin \angle K=\sin 30^\circ=\sin(90 - 60)$. The side opposite $\angle K$ is $JL = 3$, so by Sine Law: $\frac{KL}{\sin 90^\circ}=\frac{JL}{\sin \angle K}$, so $KL=\frac{JL}{\sin \angle K}=\frac{3}{\sin(90 - 60)}$ (which is option C? But earlier we had option A: $\frac{3}{\sin 60}$. Wait, there's a confusion. Wait $\angle L = 60^\circ$, side opposite is $JK = 5.2$; $\angle K = 30^\circ$, side opposite is $JL = 3$; hypotenuse $KL$. So $\sin \angle L=\frac{JK}{KL}\implies KL=\frac{JK}{\sin \angle L}=\frac{5.2}{\sin 60^\circ}$? No, wait $\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}}$. In right triangle, $\sin L=\frac{JK}{KL}\implies KL=\frac{JK}{\sin L}=\frac{5.2}{\sin 60^\circ}$? But that's not matching. Wait, no: in right triangle at $J$, so $\sin L=\frac{JK}{KL}$ (opposite over hypotenuse), so $KL=\frac{JK}{\sin L}=\frac{5.2}{\sin 60^\circ}$? But also $\cos L=\frac{JL}{KL}\implies KL=\frac{JL}{\cos L}$. And $\cos L=\sin(90 - L)=\sin(30^\circ)$, so $KL=\frac{JL}{\sin(30^\circ)}=\frac{3}{\sin(30^\circ)}=\frac{3}{\sin(90 - 60)}$. Wait, now I see the mistake: $\sin(90 - 60)=\sin 30=\cos 60$, and $\cos 60=\frac{JL}{KL}\implies KL=\frac{JL}{\cos 60}=\frac{3}{\sin(90 - 60)}$ (since $\cos 60=\sin 30$). And $\sin 60=\frac{JK}{KL}\implies KL=\frac{JK}{\sin 60}=\frac{5.2}{\sin 60}$? No, this is conflicting. Wait, let's use the right - triangle definitions:

In right triangle $JKL$, right - angled at $J$:

- $\sin L=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{JK}{KL}\implies KL = \frac{JK}{\sin L}=\frac{5.2}{\sin 60^\circ}$ (but this is not option B? Wait option B is $\frac{5.2}{\sin 60}$, yes. And $\cos L=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{JL}{KL}\implies KL=\frac{JL}{\cos L}$. Since $\cos L=\sin(90 - L)=\sin(30^\circ)$, so $KL=\frac{JL}{\sin(30^\circ)}=\frac{3}{\sin(90 - 60^\circ)}$ (option C? No, option C is $\frac{3}{\sin(90 - 60)}$, and option A is $\frac{3}{\sin 60}$. Wait, $\sin 60=\cos 30$, $\sin 30=\cos 60$. Let's calculate the values:

$\sin 60^\circ=\frac{\sqrt{3}}{2}\approx0.866$, $\sin 30^\circ = 0.5$.

Option A: $\frac{3}{\sin 60}\approx\frac{3}{0.866}\approx3.464$

Option B: $\frac{5.2}{\sin 60}\approx\frac{5.2}{0.866}\approx6.00$

Option C: $\frac{3}{\sin 30}=\frac{3}{0.5}=6$

Option D: $\frac{5.2}{\sin 30}=\frac{5.2}{0.5}=10.4$

Now, let's find the length of $KL$ using Pythagoras: $KL=\sqrt{3^2 + 5.2^2}=\sqrt{9 + 27.04}=\sqrt{36.04}\approx6.003$. So $KL\approx6$. Now, option A: $\frac{3}{\sin 60}\approx3.464$ (wrong), option B: $\frac{5.2}{\sin 60}\approx\frac{5.2}{0.866}\approx6$ (correct), option C: $\frac{3}{\sin 30}=6$ (correct), option D: $\frac{5.2}{\sin 30}=10.4$ (wrong). Wait, so earlier mistake in Sine Law application. Let's redo Sine Law:

In $\triangle JKL$, $\angle J = 90^\circ$, $\angle L = 60^\circ$, $\angle K=30^\circ$.

Sides:

- Opposite $\angle L$ (60°): $JK = 5.2$

- Opposite $\angle K$ (30°): $JL = 3$

- Hypotenuse: $KL$

By Sine Law: $\frac{KL}{\sin 90^\circ}=\frac{JK}{\sin 60^\circ}=\frac{JL}{\sin 30^\circ}$

Since $\sin 90^\circ = 1$, $\sin 30^\circ = 0.5$, $\sin 60^\circ=\frac{\sqrt{3}}{2}$

So $\frac{KL}{1}=\frac{5.2}{\sin 60^\circ}\implies KL=\frac{5.2}{\sin 60^\circ}$ (option B)

And $\frac{KL}{1}=\frac{3}{\sin 30^\circ}\implies KL=\frac{3}{\sin 30^\circ}=\frac{3}{\sin(90^\circ - 60^\circ)}$ (option C)

Wait, now I see the earlier mistake: I mixed up the opposite sides. $\angle L = 60^\circ$ has opposite side $JK = 5.2$, not $JL$. So correct:

- $\angle L = 60^\circ$, opposite side $JK = 5.2$

- $\angle K = 30^\circ$, opposite side $JL = 3$

- Hypotenuse $KL$

So Sine Law: $\frac{KL}{\sin 90^\circ}=\frac{JK}{\sin 60^\circ}\implies KL=\frac{JK}{\sin 60^\circ}=\frac{5.2}{\sin 60^\circ}$ (option B)

And $\frac{KL}{\sin 90^\circ}=\frac{JL}{\sin 30^\circ}\implies KL=\frac{JL}{\sin 30^\circ}=\frac{3}{\sin(90^\circ - 60^\circ)}$ (since $\sin 30^\circ=\sin(90^\circ - 60^\circ)$), which is option C? Wait no, option C is $\frac{3}{\sin(90 - 60)}$, and option B is $\frac{5.2}{\sin 60}$. And when we calculate $KL$ via Pythagoras: $KL=\sqrt{3^2 + 5.2^2}=\sqrt{9 + 27.04}=\sqrt{36.04}\approx6.003$

Now, $\frac{5.2}{\sin 60^\circ}=\frac{5.2}{\frac{\sqrt{3}}{2}}=\frac{10.4}{\sqrt{3}}\approx\frac{10.4}{1.732}\approx6.00$, which matches.

$\frac{3}{\sin(90^\circ - 60^\circ)}=\frac{3}{\sin 30^\circ}=\frac{3}{0.5}=6$, which also matches.

Wait, so earlier I had the opposite sides wrong. The side opposite $\angle L$ (60°) is $JK = 5.2$, and side opposite $\angle K$ (30°) is $JL = 3$. So correct options are B and C? But the options are A: $\frac{3}{\sin 60}$, B: $\frac{5.2}{\sin 60}$, C: $\frac{3}{\sin(90 - 60)}$, D: $\frac{5.2}{\sin(90 - 60)}$

Wait, let's check the angle - side correspondence again. In right triangle at $J$, so:

- Angle at $L$: 60°, so the sides:

- Adjacent to $L$: $JL = 3$

- Opposite to $L$: $JK = 5.2$

- Hypotenuse: $KL$

So $\sin L=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{JK}{KL}\implies KL=\frac{JK}{\sin L}=\frac{5.2}{\sin 60^\circ}$ (option B)

$\cos L=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{JL}{KL}\implies KL=\frac{JL}{\cos L}$. But $\cos L=\sin(90^\circ - L)=\sin(30^\circ)$, so $KL=\frac{JL}{\sin(30^\circ)}=\frac{3}{\sin(90^\circ - 60^\circ)}$ (option C)

And $\sin(90^\circ - 60^\circ)=\sin 30^\circ = 0.5$, $\sin 60^\circ=\frac{\sqrt{3}}{2}\approx0.866$

Calculating option B: $\frac{5.2}{\sin 60^\circ}\approx\frac{5.2}{0.866}\approx6.00$

Option C: $\frac{3}{\sin 30^\circ}=\frac{3}{0.5}=6$

Which matches the Pythagorean theorem result ($\sqrt{3^2 + 5.2^2}=\sqrt{9 + 27.04}=\sqrt{36.04}\approx6.003$)

Now, what about option A: $\frac{3}{\sin 60^\circ}\approx\frac{3}{0.866}\approx3.464
eq6$, so wrong.

Option D: $\frac{5.2}{\sin 30^\circ}=\frac{5.2}{0.5}=10.4
eq6$, so wrong.

Wait, but the problem says "Choose 2 answers". So the correct answers are B and C? But earlier I thought A, but that was a mistake. Wait, no, let's check the original diagram again. The diagram shows:

- Right angle at $J$

- $JL = 3$

- $JK = 5.2$

- Angle at $L$: 60°

So angle at $K$ is $90 - 60 = 30°$

So by Sine Law:

$\frac{KL}{\sin 90°}=\frac{JK}{\sin 60°}=\frac{JL}{\sin 30°}$

Since $\sin 90° = 1$, $\sin 30° = 0.5$, $\sin 60°=\frac{\sqrt{3}}{2}$

So $KL=\frac{JK}{\sin 60°}=\frac{5.2}{\sin 60°}$ (option B)

and $KL=\frac{JL}{\sin 30°}=\frac{3}{\sin(90° - 60°)}$ (option C)

Yes, so the correct options are B and C? Wait, but let's check the options again:

Option A: $\frac{3}{\sin 60°}$

Option B: $\frac{5.2}{\sin 60°}$

Option C: $\frac{3}{\sin(90° - 60°)}$

Option D: $\frac{5.2}{\sin(90° - 60°)}$

So B and C are correct. Wait, but when I first calculated, I had a mix - up of opposite sides. The key is to correctly identify the opposite sides of the angles. $\angle L = 60°$ has opposite side $JK = 5.2$, $\angle K = 30°$ has opposite side $JL = 3$, hypotenuse $KL$. So Sine Law gives $KL=\frac{JK}{\sin 60°}$ (B) and $KL=\frac{JL}{\sin 30°}=\frac{3}{\sin(90 - 60)}$ (C)

Answer:

B. $\boldsymbol{\frac{5.2}{\sin(60^{\circ})}}$, C. $\boldsymbol{\frac{3}{\sin(90^{\circ}- 60^{\circ})}}$