Question

find m∠h.
triangle hig with right angle at i, ig = 2.5, hg = 7.5
write your answer as an integer or as a decimal rounded to the nearest tenth.
m∠h =
submit

Explanation:

Step1: Identify triangle type and trigonometric ratio

We have a right triangle \( \triangle HIG \) with right angle at \( I \). To find \( m\angle H \), we can use the tangent function, where \( \tan(\angle H)=\frac{\text{opposite}}{\text{adjacent}} \). Here, the opposite side to \( \angle H \) is \( IG = 2.5 \) and the adjacent side is \( HG = 7.5 \)? Wait, no, wait. Wait, in right triangle \( HIG \), right-angled at \( I \), so sides: \( HI \) is one leg, \( IG \) is another leg, and \( HG \) is the hypotenuse? Wait, no, the length given for \( HG \) is 7.5? Wait, no, looking at the diagram: \( H \) to \( G \) is 7.5, \( I \) to \( G \) is 2.5, and \( \angle I \) is right angle. So \( \triangle HIG \) is right-angled at \( I \), so \( IG = 2.5 \) (opposite to \( \angle H \)), and \( HG = 7.5 \)? Wait, no, \( HI \) is adjacent, \( IG \) is opposite, and \( HG \) is hypotenuse? Wait, no, \( H \) to \( G \) is the hypotenuse? Wait, no, \( H \) to \( I \) is one leg, \( I \) to \( G \) is another leg, and \( H \) to \( G \) is hypotenuse. Wait, the length of \( IG \) is 2.5, \( HG \) is 7.5? Wait, no, maybe \( HG \) is the hypotenuse? Wait, no, let's re-examine. The triangle has vertices \( H \), \( I \), \( G \), right-angled at \( I \). So sides: \( HI \) (horizontal), \( IG \) (vertical), \( HG \) (hypotenuse). Wait, the length of \( IG \) is 2.5, and \( HG \) is 7.5? Wait, no, maybe \( HG \) is the hypotenuse? Wait, no, the problem says \( H \) to \( G \) is 7.5, \( I \) to \( G \) is 2.5. So in right triangle \( HIG \), right-angled at \( I \), we can use sine, cosine, or tangent. Let's use sine: \( \sin(\angle H)=\frac{IG}{HG} \), because \( IG \) is opposite to \( \angle H \), and \( HG \) is hypotenuse. Wait, \( IG = 2.5 \), \( HG = 7.5 \). So \( \sin(\angle H)=\frac{2.5}{7.5}=\frac{1}{3} \). Then \( \angle H=\arcsin(\frac{1}{3}) \). Alternatively, maybe \( HG \) is the adjacent side? Wait, no, \( H \) to \( G \) is the hypotenuse. Wait, maybe I made a mistake. Wait, the side \( HI \) is adjacent to \( \angle H \), \( IG \) is opposite, and \( HG \) is hypotenuse. So \( \sin(\angle H)=\frac{IG}{HG}=\frac{2.5}{7.5}=\frac{1}{3} \). Let's calculate \( \arcsin(\frac{1}{3}) \).

Step2: Calculate the angle

Using a calculator, \( \arcsin(\frac{1}{3}) \approx 19.47^\circ \), which rounds to \( 19.5^\circ \) when rounded to the nearest tenth. Wait, but let's check with tangent. Wait, maybe \( HG \) is the adjacent side? Wait, no, if \( \angle I \) is right angle, then \( HI \) and \( IG \) are legs, \( HG \) is hypotenuse. Wait, maybe the length of \( HI \) is 7.5? Wait, the diagram shows \( H \) to \( G \) as 7.5, \( I \) to \( G \) as 2.5. So \( HI \) is the horizontal leg, \( IG \) is vertical leg, \( HG \) is hypotenuse. So \( HI \) can be calculated by Pythagoras: \( HI=\sqrt{HG^2 - IG^2}=\sqrt{7.5^2 - 2.5^2}=\sqrt{56.25 - 6.25}=\sqrt{50}\approx7.071 \). But then \( \tan(\angle H)=\frac{IG}{HI}=\frac{2.5}{\sqrt{50}}\approx\frac{2.5}{7.071}\approx0.3535 \), and \( \arctan(0.3535)\approx19.47^\circ \), same as before. So regardless, the angle is approximately \( 19.5^\circ \) when rounded to the nearest tenth.

Answer:

\( 19.5 \)