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=(1)/(2).`. Also, hypotenuse h = 7 inches. We are to find the length of the shortest side of ΔABC.

We have,

`\cos \angle ACB=(BC)/(AC)\\\\\\\Rightarrow (1)/(2)=(b)/(h)\\\\\\\Rightarrow (1)/(2)=(b)/(7)\\\\\\\Rightarrow b=(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((7)/(2)\right)^2\\\\\\\Rightarrow p^2=(3 * 49)/(4)\\\\\\\Rightarrow p=\sqrt 3*(7)/(2)=\sqrt 3* 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.