Use the Pythagorean theorem, r^2 = y^2 + x^2
r^2
= 4^2 + 11^2
= 16 + 121
= 137
= Square root of 137
= 11.705
I need help with this:
Given a right triangle with B=4 and
C=11 find A appoximated to three decimal places.
2 answers
This is how I would do it.
Pythagorean theorem: a2 + b2 = c2.
a^2 + 4^2 = 11^2
a^2 +16 = 105
subtract both sides by 16
a^2 - 16 = 105 - 16
a^2 = 105
a = sqrt 105
a = 10.247
Pythagorean theorem: a2 + b2 = c2.
a^2 + 4^2 = 11^2
a^2 +16 = 105
subtract both sides by 16
a^2 - 16 = 105 - 16
a^2 = 105
a = sqrt 105
a = 10.247