Question

solve the right triangle shown in the figure. b = 6, c = 8 a = (round to the nearest hundredth as needed.) a = ° (round to the nearest tenth as needed.) b = ° (round to the nearest tenth as needed.)

Explanation:

Step1: Find side a using Pythagorean theorem

By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, so $a = \sqrt{c^{2}-b^{2}}$. Substituting $b = 6$ and $c = 8$, we have $a=\sqrt{8^{2}-6^{2}}=\sqrt{64 - 36}=\sqrt{28}\approx5.29$.

Step2: Find angle A using cosine function

$\cos(A)=\frac{b}{c}$. So $A=\cos^{-1}(\frac{b}{c})$. Substituting $b = 6$ and $c = 8$, we get $A=\cos^{-1}(\frac{6}{8})=\cos^{-1}(0.75)\approx41.4^{\circ}$.

Step3: Find angle B

Since the sum of angles in a triangle is $180^{\circ}$ and this is a right - triangle with $\angle C = 90^{\circ}$, then $B=90^{\circ}-A$. So $B = 90^{\circ}-41.4^{\circ}=48.6^{\circ}$.

Answer:

$a\approx5.29$
$A\approx41.4^{\circ}$
$B\approx48.6^{\circ}$