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Given the following triangle, if c = sqrt 5 and a = 2, find the measure of B to the nearest degree.

Given the following triangle, if c = sqrt 5 and a = 2, find the measure of B to the-example-1

2 Answers

7 votes

Answer: Hi!

For a triangle rectangle the length of the hypotenuse c is equal to
\sqrt{a^(2)+  b^(2) } where a and b are the length of both cathetus.

So here we have c =
√(5) and a = 2.

then
a^(2) + b^(2) = 5

4 +
b^(2) = 5

and b = 1.

If we want the angle B. then Tg(B)=
(b)/(a) = 1/2

so B = aTg(0.5) = 26° aprox.

User David Duran
by
7.0k points
4 votes
The measure of B will be given by:
cos θ= adjacent/hypotenuse
θ=B
adjacent=a=2
hypotenuse=c=√5
thus;
cos B=2/√5
thus
B=arccos 2/√5
B=26.565°~26°
User Pawka
by
6.5k points
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