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Question
which shows the correct trig equation? cosθ=8/15 sinθ=8/15 tanθ=15/8 sinθ=15/8
Step1: Recall trigonometric ratios
In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
For the given right - triangle with angle \(\theta\), the side opposite to \(\theta\) is \(8\), the hypotenuse is \(15\).
Step2: Calculate \(\sin\theta\)
Using the formula \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), we substitute the values. So \(\sin\theta=\frac{8}{15}\).
For \(\cos\theta\), the adjacent side to \(\theta\) (using Pythagoras' theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 15\) and \(b = 8\), then \(a=\sqrt{15^{2}-8^{2}}=\sqrt{225 - 64}=\sqrt{161}
eq8\)). For \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{8}{\sqrt{161}}
eq\frac{15}{8}\). Also, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\frac{15}{8}>1\) (since hypotenuse \(>\) opposite in a right - triangle, \(\sin\theta\leq1\)) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}
eq\frac{8}{15}\) (as shown by side lengths).
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\(\sin\theta=\frac{8}{15}\) (the blue option)