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]2 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]2 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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