Question

solve the right triangle using the given information.
a = 75.3 yd, b = 40.2 yd

c = 85.4 yd
(simplify your answer. type an integer or a decimal. round to the nearest tenth if needed.)
a = \\(\square^{\circ}\square\\)
(simplify your answers. type integers. round to the nearest ten minutes if needed.)
b = \\(\square^{\circ}\square\\)
(simplify your answers. type integers. round to the nearest ten minutes if needed.)

Explanation:

Step1: Find angle A using tangent

In a right triangle, $\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b}$. So $\tan(A) = \frac{75.3}{40.2}$. Calculate this value: $\frac{75.3}{40.2} \approx 1.8731$. Then find the arctangent: $A = \arctan(1.8731) \approx 61.9^\circ$. Convert the decimal part to minutes: $0.9^\circ \times 60 = 54'$, so $A \approx 61^\circ 50'$ (rounded to nearest ten minutes, 54' is closer to 50'? Wait, no, 0.9*60=54, so 61°54', but maybe my initial calculation. Wait, let's recalculate $\arctan(75.3/40.2)$. Let's compute 75.3 ÷ 40.2: 75.3 / 40.2 = 1.873134328. $\arctan(1.8731) \approx 61.9$ degrees. 0.9 degrees is 0.9*60=54 minutes, so 61°54', but maybe the problem expects rounding to nearest ten minutes. 54 is closer to 50 or 60? 54 - 50 = 4, 60 - 54 = 6, so closer to 50? Wait, no, 54 is 50 + 4, 60 - 54 = 6, so maybe 61°50'? Wait, maybe I made a mistake. Alternatively, use $\tan(A) = a/b = 75.3/40.2$. Let's check with calculator: $\arctan(75.3/40.2) \approx 61.9$ degrees. So 0.9 degrees is 54 minutes, so 61°54', but let's see angle B: since it's a right triangle, A + B = 90°, so B = 90° - A.

Step2: Find angle B

Since A + B = 90°, B = 90° - A. If A ≈ 61.9°, then B ≈ 28.1°. Convert 0.1° to minutes: 0.1*60=6', so 28°6', but rounded to nearest ten minutes, 6' is closer to 10'? Wait, no, 6 is less than 10, so maybe 28°10'? Wait, maybe my approach is wrong. Let's use $\tan(B) = b/a = 40.2/75.3 ≈ 0.5339$. Then $\arctan(0.5339) ≈ 28.1°$, which is 28°6', so rounded to nearest ten minutes, 28°10'? Wait, the problem says "round to the nearest ten minutes". So 54' is 50' or 60'? 54 is 5*10 + 4, so nearest ten is 50 or 60? 54 - 50 = 4, 60 - 54 = 6, so 50 is closer? Wait, no, 54 is 54 minutes, which is 5 tens (50) and 4, or 6 tens (60) minus 6. So 54 is closer to 50? Wait, 54 - 50 = 4, 60 - 54 = 6, so 50 is closer. So A ≈ 61°50'? Wait, but 0.9 degrees is 54 minutes, which is 54, so maybe the problem expects using calculator more accurately. Let's recalculate:

First, $a = 75.3$, $b = 40.2$, right triangle, right angle at C (since a is adjacent to B, b is adjacent to A, c is hypotenuse).

So $\tan(A) = \frac{a}{b} = \frac{75.3}{40.2} \approx 1.873134328$

$A = \arctan(1.873134328) \approx 61.903^\circ$

Convert decimal degrees to minutes: 0.903° * 60 = 54.18', so 54' (rounded to nearest minute), but the problem says "round to the nearest ten minutes". So 54' is between 50' and 60'. 54 - 50 = 4, 60 - 54 = 6, so 50' is closer? Wait, no, 54 is 5*10 + 4, so the nearest ten is 50 or 60. 54 is 4 more than 50, 6 less than 60, so 50 is closer. So A ≈ 61°50'?

Then B = 90° - 61°50' = 28°10' (since 90° is 89°60', 89°60' - 61°50' = 28°10'). Let's check with $\tan(B) = b/a = 40.2/75.3 ≈ 0.5339$, $\arctan(0.5339) ≈ 28.103°$, which is 28°6.18', so rounded to nearest ten minutes, 10' (since 6.18 is closer to 10 than 0? Wait, no, 6.18 is between 0 and 10, so nearest ten is 10? Wait, 28°6' is closer to 28°10' than 28°0'? 6 - 0 = 6, 10 - 6 = 4, so yes, 10' is closer. So B ≈ 28°10'.

Wait, maybe the initial calculation of A: 61.903° is 61°54', which is 54', so nearest ten minutes: 54 is 5*10 + 4, so 50 or 60. 54 - 50 = 4, 60 - 54 = 6, so 50 is closer? But 54 is 4 away from 50, 6 away from 60, so 50 is closer. So A = 61°50', B = 28°10'.

Let's verify with Pythagoras: $c = \sqrt{a^2 + b^2} = \sqrt{75.3^2 + 40.2^2} = \sqrt{5670.09 + 1616.04} = \sqrt{7286.13} ≈ 85.36 ≈ 85.4$, which matches the given c.

So:

A: $\arctan(75.3/40.2) ≈ 61.9°$ → 61°54' → rounded to nearest ten minutes: 61°50' (wait, no, 54 is closer to 50? Wait, 54 minutes is 54, so 50 is 5*10, 60 is 6*10. 54 - 50 = 4, 60 - 54 = 6, so 50 is closer. So A = 61°50'.

B: 90° - 61°50' = 28°10'.

Alternatively, maybe the problem expects using calculator for degrees and then converting to minutes with rounding to nearest ten. Let's do:

For A:

$\tan(A) = 75.3 / 40.2 ≈ 1.8731$

$A = \arctan(1.8731) ≈ 61.903°$

Decimal part: 0.903 * 60 = 54.18 minutes, so 54 minutes. Rounded to nearest ten minutes: 54 is between 50 and 60. The difference from 50 is 4, from 60 is 6, so 50 is closer. So A = 61°50'.

For B:

$B = 90° - 61°50' = 28°10'$ (since 90° = 89°60', 89°60' - 61°50' = 28°10').

Let's check with $\tan(B) = 40.2 / 75.3 ≈ 0.5339$

$B = \arctan(0.5339) ≈ 28.103°$

Decimal part: 0.103 * 60 ≈ 6.18 minutes, so 6 minutes. Rounded to nearest ten minutes: 6 is between 0 and 10. Difference from 0 is 6, from 10 is 4, so 10 is closer. So B = 28°10'.

Yes, that makes sense.

Answer:

$c = 85.4$ yd
$A = 61^\circ 50'$
$B = 28^\circ 10'$