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I need help with this It’s from my trig prep book

I need help with this It’s from my trig prep book-example-1
User Sanzante
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1 Answer

5 votes
5 votes

Let's draw a picture of our triangle:

We know that the larger acute angle is always in front of the large leg (denoted by x). So, we can relate the unknown angle with the shorter leg and the hypotenuse by means of the cosine function, that is,


\cos \theta=\frac{\text{ adjancent leg to theta}}{\text{hypotenuse}}=\frac{2\sqrt[]{6}}{2\sqrt[]{15}}

which gives


\begin{gathered} \cos \theta=\frac{\sqrt[]{6}}{\sqrt[]{15}}=\frac{\sqrt[]{2*3}}{\sqrt[]{5*3}} \\ \cos \theta=\frac{\sqrt[]{2}}{\sqrt[]{5}} \end{gathered}

Then, we have


\cos \theta=0.632455

Now, by applying the respective inverse function, we have


\theta=\cos ^(-1)(0.632455)

which gives


\theta=50.768

Therefore, by rounding the nearest tenth, the answer is 50.8 degrees

I need help with this It’s from my trig prep book-example-1
User Myk Willis
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