Question

use the diagram to complete the statements. the measure of angle l is 70°. the trigonometric ratio that uses ∠m and ln to solve for nm is tangent. the length of nm, to the nearest tenth, is approximately dropdown with 7.6, 22.3, 57.7. diagram: right triangle lnm with right angle at n, ln = 21, angle at m is 20°.

Explanation:

Step1: Identify the trigonometric ratio

We know that \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). For \(\angle M = 20^\circ\), the opposite side to \(\angle M\) is \(LN = 21\) and the adjacent side is \(NM\) (let's call it \(x\)). So \(\tan(20^\circ)=\frac{LN}{NM}\).

Step2: Solve for \(NM\)

From \(\tan(20^\circ)=\frac{21}{x}\), we can rearrange to get \(x=\frac{21}{\tan(20^\circ)}\). We know that \(\tan(20^\circ)\approx0.3640\). Then \(x = \frac{21}{0.3640}\approx57.7\). Wait, no, wait. Wait, angle at \(L\) is \(70^\circ\), angle at \(M\) is \(20^\circ\). Wait, maybe I mixed up. Wait, in right triangle \(LNM\), right - angled at \(N\). So \(\angle L = 70^\circ\), \(\angle M=20^\circ\), \(LN = 21\) (opposite to \(\angle M\)), \(NM\) is adjacent to \(\angle M\), \(LM\) is hypotenuse. Wait, if we use \(\angle M\), \(\tan(\angle M)=\frac{LN}{NM}\), so \(NM=\frac{LN}{\tan(\angle M)}\). \(\tan(20^\circ)\approx0.3640\), \(LN = 21\), so \(NM=\frac{21}{0.3640}\approx57.7\)? Wait, no, that can't be. Wait, maybe I got the angle wrong. Wait, \(\angle L = 70^\circ\), so \(\tan(\angle L)=\frac{NM}{LN}\). Because in right triangle \(LNM\), \(\tan(\angle L)=\frac{\text{opposite}}{\text{adjacent}}=\frac{NM}{LN}\). So \(\tan(70^\circ)=\frac{NM}{21}\). Then \(NM = 21\times\tan(70^\circ)\). \(\tan(70^\circ)\approx2.7475\), so \(NM=21\times2.7475 = 57.6975\approx57.7\).

Answer:

57.7