Question

consider △lmn. m∠l + m∠m = 90° sin(l) = sin(m) = cos(l) cos(m) cos(n)

Explanation:

Step1: Identify triangle type

$\triangle LMN$ is right - angled at $N$, so $\angle L+\angle M = 90^{\circ}$ (since the sum of angles in a triangle is $180^{\circ}$ and $\angle N = 90^{\circ}$). Given $\angle L=42^{\circ}$ and $\angle M = 48^{\circ}$.

Step2: Recall co - function identity

We know the co - function identity: $\sin\theta=\cos(90^{\circ}-\theta)$. Since $\angle L+\angle M = 90^{\circ}$, then $\angle M=90^{\circ}-\angle L$ and $\angle L = 90^{\circ}-\angle M$.
For $\sin(L)$: Since $\angle M = 90^{\circ}-\angle L$, then $\sin(L)=\cos(M)$? Wait, no, let's correct. Wait, $\angle L = 42^{\circ}$, $\angle M=48^{\circ}$, and $90^{\circ}-\angle L=\angle M$, $90^{\circ}-\angle M=\angle L$.
The co - function identity is $\sin\alpha=\cos(90^{\circ}-\alpha)$. So $\sin(L)=\sin(42^{\circ})=\cos(90 - 42)^{\circ}=\cos(48^{\circ})=\cos(M)$? Wait, no, the question is about $\sin(M)$. Let's re - evaluate.
Since $\angle M = 48^{\circ}$ and $\angle L=42^{\circ}$, and $\angle L+\angle M = 90^{\circ}$, so $\angle M=90^{\circ}-\angle L$. Then $\sin(M)=\sin(90^{\circ}-\angle L)=\cos(L)$ (by co - function identity $\sin(90^{\circ}-x)=\cos(x)$).
Wait, let's check with angles. $\sin(48^{\circ})=\cos(42^{\circ})$ because $90 - 48 = 42$. And $\angle L = 42^{\circ}$, so $\sin(M)=\cos(L)$.
Wait, the first blank for $\sin(L)$: $\angle L = 42^{\circ}$, $\angle M = 48^{\circ}$, and $\sin(L)=\sin(42^{\circ})=\cos(48^{\circ})=\cos(M)$? No, wait the options for $\sin(L)$ and $\sin(M)$ are from the dropdown which has $\cos(L)$, $\cos(M)$, $\cos(N)$. Since $\angle N = 90^{\circ}$, $\cos(N)=\cos(90^{\circ}) = 0$.
For $\sin(L)$: $\angle L = 42^{\circ}$, $\sin(42^{\circ})=\cos(48^{\circ})=\cos(M)$? Wait, no, the angle complementary to $\angle L$ is $\angle M$, so $\sin(L)=\cos(M)$? Wait, no, $\sin(\theta)=\cos(90^{\circ}-\theta)$. So if $\theta = L$, then $90^{\circ}-\theta=M$, so $\sin(L)=\cos(M)$. If $\theta = M$, then $90^{\circ}-\theta = L$, so $\sin(M)=\cos(L)$.
Looking at the triangle, $\angle N = 90^{\circ}$, so $\cos(N)=\cos(90^{\circ}) = 0$, which is not equal to $\sin(L)$ or $\sin(M)$ in this case.
So for $\sin(L)$: Since $\angle L = 42^{\circ}$, $\sin(L)=\cos(M)$ (because $90 - 42=48=\angle M$). For $\sin(M)$: $\sin(M)=\cos(L)$ (because $90 - 48 = 42=\angle L$). But the first dropdown (for $\sin(L)$) and the second (for $\sin(M)$). Wait, the problem shows that $m\angle L + m\angle M=90^{\circ}$ is already filled as $90^{\circ}$. Now, for $\sin(L)$: Let's see, $\angle L = 42^{\circ}$, $\sin(42^{\circ})=\cos(48^{\circ})=\cos(M)$? Wait, no, the options for $\sin(L)$ are in the dropdown. Wait, the dropdown has $\cos(L)$, $\cos(M)$, $\cos(N)$.
Wait, maybe I made a mistake. Let's start over.
In a right - triangle, $\sin(\text{angle})=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos(\text{angle})=\frac{\text{adjacent}}{\text{hypotenuse}}$.
For $\angle L$: opposite side is $MN$, adjacent side is $LN$, hypotenuse is $LM$.
For $\angle M$: opposite side is $LN$, adjacent side is $MN$, hypotenuse is $LM$.
$\sin(L)=\frac{MN}{LM}$, $\cos(M)=\frac{MN}{LM}$ (because for $\angle M$, adjacent side is $MN$). So $\sin(L)=\cos(M)$.
$\sin(M)=\frac{LN}{LM}$, $\cos(L)=\frac{LN}{LM}$ (because for $\angle L$, adjacent side is $LN$). So $\sin(M)=\cos(L)$.
Since $\angle N = 90^{\circ}$, $\cos(N)=0$, so it's not a valid option for $\sin(L)$ or $\sin(M)$.
So for $\sin(L)$, the value is $\cos(M)$? Wait, no, the dropdown options are $\cos(L)$, $\cos(M)$, $\cos(N)$. Wait, maybe the first blank (for $\sin(L)$) is $\cos(M)$? No, wait the angle of $\angle L$ is $42^{\circ}$, $\angle M$ is $48^{\circ}$. $\sin(42^{\circ})=\cos(48^{\circ})=\cos(M)$, and $\sin(48^{\circ})=\cos(42^{\circ})=\cos(L)$.
So for $\sin(M)$: $\sin(M)=\cos(L)$ (because $M = 48^{\circ}$, $L = 42^{\circ}$, and $\sin(48^{\circ})=\cos(42^{\circ})$).

Answer:

For $\sin(L)$: $\cos(M)$ (but wait, the options are $\cos(L)$, $\cos(M)$, $\cos(N)$). Wait, no, let's check the triangle again. $\angle L = 42^{\circ}$, so $\sin(L)=\sin(42^{\circ})=\cos(90 - 42)^{\circ}=\cos(48^{\circ})=\cos(M)$. For $\sin(M)=\sin(48^{\circ})=\cos(90 - 48)^{\circ}=\cos(42^{\circ})=\cos(L)$. So if the question is about $\sin(M)$, the answer is $\cos(L)$. But the first blank (maybe I misread). Wait, the first dropdown is for $\sin(L)$, the second for $\sin(M)$. Let's assume the question is about $\sin(M)$ (since the dropdown is open for $\sin(M)$). So $\sin(M)=\cos(L)$.

Wait, the original problem: "sin(L) = $\sin(M) = $ with options cos(L), cos(M), cos(N)". Let's re - express:

Since $\triangle LMN$ is right - angled at $N$, $\angle L+\angle M = 90^{\circ}$.

By co - function identity: $\sin\theta=\cos(90^{\circ}-\theta)$

For $\sin(L)$: $\theta = L$, so $\sin(L)=\cos(90^{\circ}-L)=\cos(M)$ (because $90^{\circ}-L = M$)

For $\sin(M)$: $\theta = M$, so $\sin(M)=\cos(90^{\circ}-M)=\cos(L)$ (because $90^{\circ}-M = L$)

Since $\angle N = 90^{\circ}$, $\cos(N)=\cos(90^{\circ}) = 0$, so it's not a valid option for $\sin(L)$ or $\sin(M)$.

So if we are to fill $\sin(M)$, the answer is $\cos(L)$.