QUESTION IMAGE
Question
find the missing side. round to the nearest tenth.
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Step1: Recall the cosine formula
For a right - triangle, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$
Problem 1
- $\cos72^{\circ}=\frac{6}{x}$
- $x = \frac{6}{\cos72^{\circ}}$
- $\cos72^{\circ}\approx0.3090$
- $x=\frac{6}{0.3090}\approx19.4$
Problem 2
- $\cos73^{\circ}=\frac{x}{6}$
- $x = 6\times\cos73^{\circ}$
- $\cos73^{\circ}\approx0.2924$
- $x = 6\times0.2924\approx1.8$
Problem 3
- $\tan24^{\circ}=\frac{x}{12}$
- $x = 12\times\tan24^{\circ}$
- $\tan24^{\circ}\approx0.4452$
- $x = 12\times0.4452\approx5.3$
Problem 4
- $\tan37^{\circ}=\frac{x}{12}$
- $\tan37^{\circ}\approx0.7536$
- $x = 12\times0.7536\approx9.0$
Problem 5
- $\tan49^{\circ}=\frac{14}{x}$
- $x=\frac{14}{\tan49^{\circ}}$
- $\tan49^{\circ}\approx1.1504$
- $x=\frac{14}{1.1504}\approx12.2$
Problem 6
- $\cos51^{\circ}=\frac{14}{x}$
- $x=\frac{14}{\cos51^{\circ}}$
- $\cos51^{\circ}\approx0.6293$
- $x=\frac{14}{0.6293}\approx22.2$
Problem 7
- $\cos63^{\circ}=\frac{16}{x}$
- $x=\frac{16}{\cos63^{\circ}}$
- $\cos63^{\circ}\approx0.4540$
- $x=\frac{16}{0.4540}\approx35.2$
Problem 8
- $\sin15^{\circ}=\frac{16}{x}$
- $x=\frac{16}{\sin15^{\circ}}$
- $\sin15^{\circ}\approx0.2588$
- $x=\frac{16}{0.2588}\approx62.6$
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- $19.4$
- $1.8$
- $5.3$
- $9.0$
- $12.2$
- $22.2$
- $35.2$
- $62.6$