Question

question 4, 4.8.7 points: 0 of 1 solve the right triangle shown in the figure. a = 28.5, c = 54.6 a≈□° (round to the nearest tenth as needed.) b≈□° (round to the nearest tenth as needed.) b≈□ (round to the nearest hundredth as needed.)

Explanation:

Step1: Find side b using Pythagorean theorem

By the Pythagorean theorem $b = \sqrt{c^{2}-a^{2}}$. Substitute $a = 28.5$ and $c = 54.6$ into the formula: $b=\sqrt{54.6^{2}-28.5^{2}}=\sqrt{(54.6 + 28.5)(54.6-28.5)}=\sqrt{83.1\times26.1}=\sqrt{2168.91}\approx46.57$.

Step2: Find angle A using sine function

We know that $\sin A=\frac{a}{c}$. So $A=\sin^{-1}(\frac{a}{c})$. Substitute $a = 28.5$ and $c = 54.6$: $A=\sin^{-1}(\frac{28.5}{54.6})\approx31.4^{\circ}$ (rounded to the nearest tenth).

Step3: Find angle B

Since the sum of angles in a triangle is $180^{\circ}$ and this is a right - triangle ($C = 90^{\circ}$), then $B=90^{\circ}-A$. So $B = 90^{\circ}-31.4^{\circ}=58.6^{\circ}$.

Answer:

$A\approx31.4^{\circ}$
$B\approx58.6^{\circ}$
$b\approx46.57$