Question

consider △lmn.
m∠l + m∠m = \boxed{\quad}^\circ
sin(l) = \boxed{\quad}
sin(m) = \boxed{\quad}
(image shows right triangle lmn with right angle at n, angle at l is 42°, angle at m is 48°)

Explanation:

Step1: Identify triangle type

△LMN is a right - triangle with ∠N = 90°, ∠L = 42°, ∠M = 48°.

Step2: Calculate m∠L + m∠M

We know that ∠L = 42° and ∠M = 48°. So, m∠L + m∠M=42° + 48° = 90°.

Step3: Calculate sin(L)

In a right - triangle, sin(angle)=opposite/hypotenuse. For ∠L, the opposite side to ∠L is MN and the hypotenuse is LM. Also, we know that in a right - triangle, sin(θ)=cos(90° - θ). Since ∠L = 42°, 90°-42° = 48°=∠M. So, sin(L)=sin(42°) and also, sin(42°)=cos(48°). But we can also use the fact that in right - triangle, sin(L)=\frac{MN}{LM} and sin(M)=\frac{LN}{LM}. And we know that ∠L + ∠M = 90°, so sin(L)=cos(M) and sin(M)=cos(L). So, sin(L)=sin(42°)≈0.6691, sin(M)=sin(48°)≈0.7431. But from the angle sum, we first found that m∠L + m∠M = 90°.

Answer:

For \(m\angle L + m\angle M\), the value is \(\boldsymbol{90}\).