Question

In a right triangle the cosine of an acute angle is 1/2 and the hypotenuse measures 7 inches. What is the length of the shortest side of the triangle?

Answer 1

If the cosine is 1/2, and the angle is acute, then the angle is 60 degrees. Using the rules for a 30-60-90 right triangle, hyp=2 times the short leg. So 7/2= 3.5.

Answer 2

Answer:  The answer is 3.5 inches.

Step-by-step explanation:  As given in the question and shown in the attached figure that ΔAB is right angled at ∠B = 90° and let $\cos \angle ACB=\dfrac{1}{2}.$. Also, hypotenuse h = 7 inches. We are to find the length of the shortest side of ΔABC.

We have,

$\cos \angle ACB=\dfrac{BC}{AC}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{b}{h}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{b}{7}\\\\\\\Rightarrow b=\dfrac{7}{2}=3.5.$

Now, using the Pythagoras theorem, we have

$b^2+p^2=h^2\\\\\Rightarrow p^2=h^2-b^2=7^2-\left(\dfrac{7}{2}\right)^2\\\\\\\Rightarrow p^2=\dfrac{3 \times 49}{4}\\\\\\\Rightarrow p=\sqrt 3\times\dfrac{7}{2}=\sqrt 3\times 3.5.$

since, √3 × 3.5 > 3.5, so, the shortest side of the triangle will be 3.5 inches.

Thus, the answer is 3.5 inches.