Question

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

Answer 1

Answer:

The cosine of the smallest angle of the triangle is $\frac{\sqrt{3}}{2}$

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

$AB^{2}=AC^{2}+BC^{2}$

substitute the values

$AB^{2}=x^{2}+(x\sqrt{3})^{2}$

$AB^{2}=x^{2}+3x^{2}$

$AB^{2}=4x^{2}$

$AB=2x$

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

$cos(B)=\frac{BC}{AB}$

substitute

$cos(B)=\frac{x\sqrt{3}}{2x}$

$cos(B)=\frac{\sqrt{3}}{2}$

Answer 2

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