Aright triangle has a hypotenuse length of 13 units. one leg has a length of 12 units. which equation can be used to determine the length of the other leg?
group of answer choices

latex: a^2=12^2+13^2 a 2 = 12 2 + 13 2

latex: a=\sqrt{13^2-12^2} a = 13 2 − 12 2

latex: a^2=12^2-13^2 a 2 = 12 2 − 13 2

latex: a=\sqrt{12^2+13^2} a = 12 2 + 13 2

We know that a²+b²=c²

So if you follow the pythagorean theorem,  we substitute the hypotenuse for c² and the leg (12) for either a² or b² because a and b are both legs, so it doesn't matter.

now we have a²+12²=13²  

But I usually take the shortcut..and instead of a²+12²=13²

I subtract 12² from 13² (it's the same thing)

After you subtract 13²-12²=a²...you find the square root of 13²-12²

so the answer is LaTeX: a=\sqrt{13^2-12^2}

a= square root of 13²-12²

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°