Question

1. find m∠h.

f
4√7
g
4√21
h
m∠h = °
submit

Explanation:

Step1: Identify triangle type

Triangle \( FHG \) is right - angled at \( F \). So, we can use trigonometric ratios. Let's recall that for a right - triangle, \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \). In \( \triangle FHG \), for \( \angle H \), the opposite side to \( \angle H \) is \( FG = 4\sqrt{7} \) and the adjacent side is \( FH=4\sqrt{21} \). Wait, alternatively, we can simplify the ratio of the sides. Let's find \( \tan(\angle H)=\frac{FG}{FH} \). Substitute \( FG = 4\sqrt{7} \) and \( FH = 4\sqrt{21} \). Then \( \tan(\angle H)=\frac{4\sqrt{7}}{4\sqrt{21}}=\frac{\sqrt{7}}{\sqrt{21}}=\frac{\sqrt{7}}{\sqrt{7}\times\sqrt{3}}=\frac{1}{\sqrt{3}} \). Wait, no, maybe I made a mistake. Wait, or maybe it's a 30 - 60 - 90 triangle? Wait, let's check the ratio again. Wait, \( \frac{FG}{FH}=\frac{4\sqrt{7}}{4\sqrt{21}}=\frac{1}{\sqrt{3}} \), so \( \tan(\angle H)=\frac{1}{\sqrt{3}} \), which means \( \angle H = 30^{\circ} \)? Wait, no, \( \tan(30^{\circ})=\frac{1}{\sqrt{3}} \), \( \tan(60^{\circ})=\sqrt{3} \), \( \tan(45^{\circ}) = 1 \). Wait, maybe I mixed up opposite and adjacent. Wait, in right - triangle \( \triangle FHG \), right - angled at \( F \), \( \angle H \) has opposite side \( FG \) and adjacent side \( FH \). Wait, \( FG = 4\sqrt{7} \), \( FH = 4\sqrt{21} \). Let's simplify \( \frac{FG}{FH}=\frac{\sqrt{7}}{\sqrt{21}}=\frac{1}{\sqrt{3}} \), so \( \tan(\angle H)=\frac{1}{\sqrt{3}} \), so \( \angle H = 30^{\circ} \)? Wait, no, wait \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \). If \( \angle H \) is at \( H \), then the sides: \( FH \) is one leg, \( FG \) is the other leg. Wait, maybe I should use \( \tan(\angle H)=\frac{FG}{FH} \). Let's compute \( \frac{4\sqrt{7}}{4\sqrt{21}}=\frac{\sqrt{7}}{\sqrt{21}}=\frac{1}{\sqrt{3}}\approx0.577 \), and \( \tan(30^{\circ})=\frac{1}{\sqrt{3}}\approx0.577 \), so \( \angle H = 30^{\circ} \)? Wait, no, wait \( \tan(30^{\circ})=\frac{1}{\sqrt{3}} \), so if \( \tan(\angle H)=\frac{1}{\sqrt{3}} \), then \( \angle H = 30^{\circ} \). Wait, but let's check again. Alternatively, maybe the triangle has sides in the ratio of \( 1:\sqrt{3}:2 \), which is a 30 - 60 - 90 triangle. The side opposite \( 30^{\circ} \) is the shortest side. Here, \( FG = 4\sqrt{7} \), \( FH = 4\sqrt{21}=4\sqrt{7}\times\sqrt{3} \), so the ratio of \( FG:FH = 1:\sqrt{3} \), so the angle opposite \( FG \) (which is \( \angle H \)) is \( 30^{\circ} \), and the angle opposite \( FH \) (which is \( \angle G \)) is \( 60^{\circ} \), and the hypotenuse \( HG \) would be \( 2\times FG=8\sqrt{7} \) (since in 30 - 60 - 90 triangle, hypotenuse is twice the shorter leg). So \( m\angle H = 30^{\circ} \). Wait, but let's re - check. \( \tan(\angle H)=\frac{FG}{FH}=\frac{4\sqrt{7}}{4\sqrt{21}}=\frac{1}{\sqrt{3}} \), so \( \angle H=\arctan(\frac{1}{\sqrt{3}})=30^{\circ} \).

Step2: Confirm the angle

Since \( \tan(30^{\circ})=\frac{1}{\sqrt{3}} \), and we found that \( \tan(\angle H)=\frac{1}{\sqrt{3}} \), we can conclude that \( m\angle H = 30^{\circ} \).

Answer:

\( 30 \)