△ FGH is a right triangle.

△ FGH is a right triangle.

I need more information — what exactly do you want to find about triangle FGH?

Below I’ll list the common things one is usually asked about a right triangle and show full solution methods (with examples) so you can tell me which applies to your problem or supply the missing data (side lengths, an angle, coordinates, etc.).

1) Subject and possible questions

  • Subject: Geometry / Right triangle trigonometry.
  • Typical asks: find a missing side, find an angle, compute area or perimeter, identify which side is the hypotenuse, use special-triangle ratios (30–60–90, 45–45–90), or work with coordinates.

2) Key formulas and facts (useful for any right triangle)

  • Pythagorean theorem: for legs $a,b$ and hypotenuse $c$: $$c^2=a^2+b^2.$$
  • Trigonometric ratios at an acute angle $\theta$: $$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}.$$
  • Area: $$\text{Area}=\frac{1}{2}\cdot(\text{leg}_1)\cdot(\text{leg}_2).$$
  • Perimeter: $$\text{Perimeter}= (\text{leg}_1)+(\text{leg}_2)+(\text{hypotenuse}).$$
  • Special ratios:
  • 45–45–90: legs equal, hypotenuse $= \text{leg}\cdot\sqrt{2}$.
  • 30–60–90: sides in ratio $1:\sqrt{3}:2$ (short leg : long leg : hypotenuse).

3) Worked examples (full steps)

Example A — find hypotenuse given two legs:

  • Given $FG=3$, $GH=4$ (legs). Hypotenuse is $FH$.
  • Step 1: Apply Pythagorean theorem: $$FH^2=FG^2+GH^2=3^2+4^2=9+16=25.$$
  • Step 2: Take square root: $$FH=\sqrt{25}=5.$$
  • Answer: $FH=5$.

Example B — find missing leg given hypotenuse and one leg:

  • Given $FH=13$ (hypotenuse), $FG=5$ (one leg). Find $GH$.
  • Step 1: $$GH^2=FH^2-FG^2=13^2-5^2=169-25=144.$$
  • Step 2: $$GH=\sqrt{144}=12.$$
  • Answer: $GH=12$.

Example C — find hypotenuse from one leg and an angle:

  • Given angle at G is $30^\circ$, and adjacent leg $FG=8$. Find $FH$.
  • Step 1: Use cosine: $$\cos 30^\circ=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{FG}{FH}.$$
  • Step 2: Solve: $$FH=\frac{FG}{\cos30^\circ}=\frac{8}{\sqrt{3}/2}=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}.$$
  • Answer: $FH=\dfrac{16\sqrt{3}}{3}$.

Example D — area with legs 6 and 7:

  • Area = $\tfrac12\cdot6\cdot7=21$.

4) If your problem involves coordinates

  • If the triangle’s vertices are at coordinates, compute vectors for two sides and use the dot product to check right angle: vectors u and v are perpendicular if $$u\cdot v=0.$$
  • Side lengths come from distance formula: between $(x_1,y_1)$ and $(x_2,y_2)$, $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$

5) What I need from you to proceed with the exact solution

  • Provide any two side lengths, or one side and one acute angle, or coordinates of the vertices, or a picture/diagram with side labels and numbers. If the image is the triangle, tell me whether FGH are labeled and which vertex is the right angle (if not evident).

One-line summary: Tell me which quantity you need (side, angle, area, perimeter, etc.) and give the known values (two numbers or one number+angle or coordinates), and I will compute it step-by-step.

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