Question

standard 14 homework
score: 5/30 answered: 4/18
question 5
find exact values for ( x ) and ( y ) in the right triangle shown above. no decimals allowed.
( x=)
( y=)
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Explanation:

Step1: Use cosine function to find \(x\)

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Given \(\theta = 60^{\circ}\), adjacent side \(a = 5\), and hypotenuse \(x\).
We know that \(\cos60^{\circ}=\frac{5}{x}\). Since \(\cos60^{\circ}=\frac{1}{2}\), we have \(\frac{1}{2}=\frac{5}{x}\). Cross - multiply gives \(x = 10\).

Step2: Use Pythagorean theorem or sine function to find \(y\)

Method 1: Pythagorean theorem

By the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 5\), \(c=x = 10\). Then \(5^{2}+y^{2}=10^{2}\). So \(y^{2}=100 - 25=75\). And \(y=\sqrt{75}=5\sqrt{3}\).

Method 2: Sine function

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Given \(\theta = 60^{\circ}\), hypotenuse \(x = 10\). Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\) and \(\sin60^{\circ}=\frac{y}{x}\), substituting \(x = 10\) gives \(y = 10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\).

Answer:

\(x = 10\), \(y = 5\sqrt{3}\)