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Snezhnost [94]
3 years ago
5

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

Mathematics
2 answers:
LiRa [457]3 years ago
8 0

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 = \sqrt{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)= \frac{b}{a} = 1/2

so B = aTg(0.5) = 26° aprox.

matrenka [14]3 years ago
6 0
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°
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