Question

find the measure of each side indicated. round to the nearest tenth.
9)
10)
solve each triangle. round answers to the nearest tenth.
11)
12)

Explanation:

Problem 9

Step1: Identify trigonometric ratio

We use tangent: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$
$\tan(68^\circ)=\frac{x}{4}$

Step2: Solve for $x$

Rearrange to isolate $x$:
$x = 4\times\tan(68^\circ)$
Calculate $\tan(68^\circ)\approx2.475$, so $x\approx4\times2.475=9.9$

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Problem 10

Step1: Identify trigonometric ratio

We use cosine: $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$
$\cos(10^\circ)=\frac{x}{2}$

Step2: Solve for $x$

Rearrange to isolate $x$:
$x = 2\times\cos(10^\circ)$
Calculate $\cos(10^\circ)\approx0.9848$, so $x\approx2\times0.9848=2.0$

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Problem 11

Step1: Find $\angle A$

Sum of angles in triangle is $180^\circ$:
$\angle A = 180^\circ - 90^\circ - 53^\circ = 37^\circ$

Step2: Find side $AC$

Use tangent for $\angle B$: $\tan(53^\circ)=\frac{AC}{6}$
$AC = 6\times\tan(53^\circ)\approx6\times1.3270=8.0$

Step3: Find hypotenuse $AB$

Use cosine for $\angle B$: $\cos(53^\circ)=\frac{6}{AB}$
$AB = \frac{6}{\cos(53^\circ)}\approx\frac{6}{0.6018}=10.0$

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Problem 12

Step1: Find $\angle A$

Sum of angles in triangle is $180^\circ$:
$\angle A = 180^\circ - 90^\circ - 25^\circ = 65^\circ$

Step2: Find side $BC$

Use sine for $\angle A$: $\sin(65^\circ)=\frac{BC}{11}$
$BC = 11\times\sin(65^\circ)\approx11\times0.9063=10.0$

Step3: Find side $AC$

Use cosine for $\angle A$: $\cos(65^\circ)=\frac{AC}{11}$
$AC = 11\times\cos(65^\circ)\approx11\times0.4226=4.6$

Answer:

1. $x\approx9.9$
2. $x\approx2.0$
3. $\angle A=37^\circ$, $AC\approx8.0$, $AB\approx10.0$
4. $\angle A=65^\circ$, $BC\approx10.0$, $AC\approx4.6$