The legs of a right triangle are lengths x and x square root 3. The cosine of the smallest angle of the triangle is _____.

AC=x, BC=x√3, AB is a hypotenuse (look at the pic below)
x√3>x, consequently ∠B<∠A and we have to find cosB, which is (x√3)/AB.
Let's find AB:
AB=√(x²+(x√3)²)=√(x²+3x²)=√(4x²)=2x
Thus, cosB is:
cosB=(x√3)/(2x)=√3/2

Answer Link

calculista calculista · Answer 2 · 2019-01-24T15:35:26+01:00

Answer:

The cosine of the smallest angle of the triangle is [tex]\frac{\sqrt{3}}{2}[/tex]

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In the right triangle ABC

Applying the Pythagoras Theorem

Find the length of the hypotenuse AB

[tex]AB^{2}=AC^{2}+BC^{2}[/tex]

substitute the values

[tex]AB^{2}=x^{2}+(x\sqrt{3})^{2}[/tex]

[tex]AB^{2}=x^{2}+3x^{2}[/tex]

[tex]AB^{2}=4x^{2}[/tex]

[tex]AB=2x[/tex]

In the right triangle ABC

The smallest angle is the angle opposite to the smallest side

therefore

the angle B is the smallest angle

Remember that

[tex]cos(B)=\frac{BC}{AB}[/tex]

substitute

[tex]cos(B)=\frac{x\sqrt{3}}{2x}[/tex]

[tex]cos(B)=\frac{\sqrt{3}}{2}[/tex]