Question

solve the right triangle abc, where c = 90°. give angles in degrees and minutes. a = 25.00 m, c = 29.00 m. b ≈ m (round to the nearest hundredth as needed.) a = ° (round to the nearest minute as needed.) b = ° (round to the nearest minute as needed.)

Explanation:

Step1: Find side b using Pythagorean theorem

By the Pythagorean theorem $b=\sqrt{c^{2}-a^{2}}$. Substitute $a = 25.00$ and $c=29.00$ into the formula:
$b=\sqrt{29.00^{2}-25.00^{2}}=\sqrt{(29 + 25)(29 - 25)}=\sqrt{54\times4}=\sqrt{216}\approx14.70$ m

Step2: Find angle A using sine function

$\sin A=\frac{a}{c}$. Substitute $a = 25.00$ and $c = 29.00$, then $A=\sin^{-1}(\frac{25}{29})$.
$A=\sin^{-1}(\frac{25}{29})\approx59.99^{\circ}$. Convert the decimal part to minutes: $0.99^{\circ}\times60\approx59'$, so $A\approx60^{\circ}0'$

Step3: Find angle B

Since the sum of angles in a triangle is $180^{\circ}$ and $C = 90^{\circ}$, $B=180^{\circ}-90^{\circ}-A$.
$B = 90^{\circ}-A$. Substituting $A\approx60^{\circ}0'$, we get $B\approx30^{\circ}0'$

Answer:

$b\approx14.70$ m
$A = 60^{\circ}0'$
$B = 30^{\circ}0'$