How do you solve special right triangles?

Answer 1

The only two 'special' right triangles I know of are 30-60-90 and 45-45-90 right triangles. What makes the first 'special' is that the side opposite the 30 degree angle is always half the length of the hypotenuse. That makes it easy to calculate the sine, cosine and tangent of its various angles. The 45-45-90 triangle has legs that are equal in length, so its sine and cosine are equal, again facilitating calculations. However, these are both solved the way you would other right triangles, using the Pythagorean Theorem.

Answer 2

the comparable way you resolve the different ones. you could continuously use Pythagorean Theorem. yet you additionally can memorize the ratios of the part lengths for the specific ones. on a 40 5-40 5-ninety (isosceles) the ratio is a million:a million:sqrt 2 meaning the climate opposite the 40 5 diploma angles are equivalent, and the hypoteneuse is the part length situations the sq. root of two. working example, 5:5: 5*sqrt(2) on a 30-60-ninety the ratio is a million:2:sqrt(3). The a million is going opposite the 30 stages, the two with the ninety stages, and the sqrt(3) with the 60 stages

Answer 3

I dont know what you mean by "special" and I don't know if you want to know the measurment of the slant but the formula is
c^2=a^2+b^2

Answer 4

RELAÇÕES MÉTRICAS NOS TRIÂNGULOS RETÂNGULOS
~~ Traçar um triângulo e usar as relações abaixo, de acordo com as medidas da figura dada e a necessidade.
a² = b² + c²
b² = a.m
c = a.n
h² = m.n
b.c = a.h

RELAÇÕES MÉTRICAS NUM TRIÂNGULO QUALQUER
a) Se  < 90º ==> a² = b² + c² - 2mc
b) Se  > 90º ==> a² = b² + c² + 2mc
Calculando "m" em função de "b" e "cosA" obtemos, em ambos os casos, o teorema dos co-senos:
a² = b² + c² - 2bc.cosA
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Answer 5

You use the Pythagorean theory. The link below explains it all really well. Check it out.

http://en.wikipedia.org/wiki/Pythagorean_theorem

Answer 6

as with any right triangles, you use the pythagorean theorem.