Question

solve the right triangle abc, where c = 90°
a = 77.2 yd, b = 41.7 yd

c = 87.7 yd
(simplify your answer. type an integer or a decimal. round to the nearest tenth as needed.)
a = □°□
(simplify your answers. type integers. round to the nearest ten minutes as needed.)
b = □°□
(simplify your answers. type integers. round to the nearest ten minutes as needed.)

Explanation:

Step1: Find angle A using tangent

We know that in a right triangle, \(\tan(A)=\frac{a}{b}\). Given \(a = 77.2\) yd and \(b = 41.7\) yd, so \(\tan(A)=\frac{77.2}{41.7}\approx1.8513\). Then \(A=\arctan(1.8513)\).
Using a calculator, \(A\approx61.7^{\circ}\). Now we convert the decimal part to minutes. The decimal part is \(0.7\), and since \(1^{\circ}=60'\), \(0.7\times60 = 42'\), so \(A\approx61^{\circ}40'\) (rounding to nearest ten minutes, 42' is closer to 40' or 50'? 42 - 40 = 2, 50 - 42 = 8, so 40'). Wait, actually, let's do it more accurately. First, calculate \(A=\arctan(\frac{77.2}{41.7})\). Let's compute \(\frac{77.2}{41.7}\approx1.8513\). \(\arctan(1.8513)\approx61.73^{\circ}\). The decimal part is \(0.73\), \(0.73\times60 = 43.8'\), which rounds to 40' or 50'? 43.8 is closer to 40'? No, 43.8 - 40 = 3.8, 50 - 43.8 = 6.2, so actually, maybe the problem expects using tangent or sine/cosine. Alternatively, use sine: \(\sin(A)=\frac{a}{c}\), \(c = 87.7\), so \(\sin(A)=\frac{77.2}{87.7}\approx0.8803\), then \(A=\arcsin(0.8803)\approx61.7^{\circ}\), same as before. So \(A\approx61^{\circ}40'\) (since 0.73*60=43.8, which is 40' when rounded to nearest ten minutes? Wait, maybe the problem wants to round to nearest ten minutes, so 43.8' is 40'? No, 40' is 40, 50' is 50. 43.8 is closer to 40? No, 43.8 - 40 = 3.8, 50 - 43.8 = 6.2, so actually, 40' is closer? Wait, maybe I made a mistake. Let's use tangent again. \(\tan(A)=\frac{77.2}{41.7}\approx1.8513\), so \(A=\arctan(1.8513)\approx61.73^{\circ}\). The decimal part is 0.73, multiply by 60: 0.73*60=43.8', which is 40' when rounded to nearest ten minutes? Wait, no, 43.8 is 40'? No, 40' is 40, 50' is 50. 43.8 is 40'? Wait, maybe the problem expects to round to nearest ten minutes, so 43.8' is 40'? Or maybe 40'? Wait, let's check angle B. Since \(A + B=90^{\circ}\), so \(B = 90^{\circ}-A\). Let's first find A in degrees with decimal, then convert.

Step2: Find angle A using arctangent

\(\tan(A)=\frac{a}{b}=\frac{77.2}{41.7}\approx1.8513\)
\(A=\arctan(1.8513)\approx61.7^{\circ}\) (decimal degrees)
To convert to degrees and minutes: \(0.7^{\circ}\times60 = 42'\), so \(A\approx61^{\circ}40'\) (since 42' is closer to 40'? Wait, 42' is 40' when rounded to nearest ten minutes? No, 40' and 50' are the ten-minute marks. 42 - 40 = 2, 50 - 42 = 8, so 40' is closer. Wait, but maybe the correct way is: 0.73 degrees (more accurately, 0.73) times 60 is 43.8 minutes, which is 40' when rounded to nearest ten minutes? No, 43.8 is 40'? No, 40' is 40, 50' is 50. 43.8 is 40'? Wait, maybe I should use the calculator's degree-minute conversion. Let's use a calculator: \(\arctan(77.2/41.7)\approx61.73^{\circ}\). 0.73 degrees * 60 = 43.8 minutes, which is 40' when rounded to the nearest ten minutes (since 43.8 is closer to 40 than 50? No, 43.8 - 40 = 3.8, 50 - 43.8 = 6.2, so 40' is closer. Wait, but maybe the problem expects 61°40' for A.

Step3: Find angle B

Since \(A + B=90^{\circ}\), \(B = 90^{\circ}-A\). If \(A\approx61.73^{\circ}\), then \(B\approx28.27^{\circ}\). Convert 0.27° to minutes: \(0.27\times60 = 16.2'\), which rounds to 20'? Wait, no, 0.27*60=16.2', which is 20'? Wait, 16.2 is closer to 20'? No, 10' and 20'? Wait, no, the problem says "round to the nearest ten minutes", so ten-minute intervals: 0', 10', 20', etc. 16.2' is closer to 20'? 16.2 - 10 = 6.2, 20 - 16.2 = 3.8, so 20' is closer. Wait, but let's do it properly. Alternatively, use \(\tan(B)=\frac{b}{a}=\frac{41.7}{77.2}\approx0.5402\), so \(B=\arctan(0.5402)\approx28.3^{\circ}\). 0.3°*60=18', which is 20'? Wait, 18' is closer to 20'? 18 - 10 = 8, 20 - 18 = 2, so 20'? Wait, maybe the correct values are:

First, confirm \(c\): \(c=\sqrt{a^{2}+b^{2}}=\sqrt{77.2^{2}+41.7^{2}}=\sqrt{5959.84 + 1738.89}=\sqrt{7698.73}\approx87.7\) yd (correct as given).

Now, angle A: \(\tan(A)=\frac{a}{b}=\frac{77.2}{41.7}\approx1.8513\), so \(A=\arctan(1.8513)\approx61.7^{\circ}\). Converting 0.7° to minutes: 0.7*60=42', which is 40' when rounded to nearest ten minutes (since 42 is closer to 40 than 50? Wait, 42 - 40 = 2, 50 - 42 = 8, so 40' is closer. So \(A\approx61^{\circ}40'\).

Angle B: \(90^{\circ}-61^{\circ}40' = 28^{\circ}20'\) (since 90° is 89°60', 89°60' - 61°40' = 28°20'). Let's check with \(\tan(B)=\frac{41.7}{77.2}\approx0.5402\), \(\arctan(0.5402)\approx28.3^{\circ}\), 0.3°*60=18', which is 20' when rounded to nearest ten minutes (18 is closer to 20 than 10). So that matches.

Answer:

\(A = 61^{\circ}40'\)
\(B = 28^{\circ}20'\)
\(c = 87.7\) yd