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use the following to solve the right triangle. round side lengths to tw…

Question

use the following to solve the right triangle. round side lengths to two decimal places
b = 65 4834°, c = 3445.3 m
image of right triangle with right angle at c, vertices a, b, c; hypotenuse c, side a (opposite a), side b (adjacent to a)
options:
○ a. a=24 5160°, a=1,348.83 m, b=3,119.38 m
○ b. a=24 5160°, a=1,429.65 m, b=3,089.29 m
○ c. a=24 5160°, a=1,429.65 m, b=3,134.68 m
○ d. a=24 5160°, a=1,485.96 m, b=3,134.68 m

Explanation:

Step1: Find angle A

In a right triangle, the sum of angles is \(180^\circ\), and \(\angle C = 90^\circ\), \(\angle B = 65.4834^\circ\). So \(\angle A=180^\circ - 90^\circ - 65.4834^\circ = 24.5166^\circ\).

Step2: Find side a (opposite to angle B)

Using sine function: \(\sin B=\frac{a}{c}\), so \(a = c\times\sin B\). Given \(c = 3445.3\) m, \(\sin(65.4834^\circ)\approx\sin(65.4834)\). Calculate \(a = 3445.3\times\sin(65.4834^\circ)\approx3445.3\times0.9093\approx3134.68\)? Wait, no, wait the options have smaller numbers. Wait, maybe I misread the diagram. Wait, the diagram: right angle at C, so side c is hypotenuse, side a is opposite angle A? Wait no, angle B: in triangle ABC, right-angled at C, so angle B is at vertex B, so side a is opposite angle A, side b opposite angle B? Wait no, standard notation: in right triangle ABC, right-angled at C, then side opposite A is a, opposite B is b, hypotenuse c. So \(\sin B=\frac{b}{c}\), \(\cos B=\frac{a}{c}\). Oh! I made a mistake. So angle B: \(\sin B=\frac{b}{c}\), \(\cos B=\frac{a}{c}\).

So let's correct:

\(\angle A = 90^\circ - 65.4834^\circ = 24.5166^\circ\) (since in right triangle, angles are complementary).

Now, side a: adjacent to angle B, so \(a = c\times\cos B\).

Side b: opposite to angle B, so \(b = c\times\sin B\).

Wait, but the options have a and b around 1k - 3k, but c is 3445.3 m. Wait, maybe the c is 3445.3? No, the options have a and b as ~1k - 3k, but 3445.3 is hypotenuse. Wait, maybe the c is 3445.3? Wait no, the options have a and b as 1,429 etc., so maybe c is 3445.3? Wait no, 3445.3 is too big. Wait, maybe the c is 34.453 m? Maybe a typo? Wait the problem says c = 3445.3 m, but options have a and b as ~1k - 3k. Wait, let's check the options.

Wait the options:

A: a=1,348.83, b=3,119.38

B: a=1,429.65, b=3,089.20

C: a=1,429.65, b=3,134.68

D: a=1,485.96, b=3,134.68

Wait, let's calculate \(\cos(65.4834^\circ)\) and \(\sin(65.4834^\circ)\).

First, convert 65°48'34'' to decimal: 48 minutes is 48/60 = 0.8 degrees, 34 seconds is 34/3600 ≈ 0.00944 degrees, so total B = 65 + 0.8 + 0.00944 ≈ 65.80944°? Wait no, the problem says B = 65 4834°, maybe 65.4834°? Let's assume B = 65.4834°.

\(\cos(65.4834^\circ)\approx\cos(65.4834)\approx0.419\) (wait no, cos(60°)=0.5, cos(65°)≈0.4226, cos(65.4834°)≈cos(65 + 0.4834°)=cos(65°)cos(0.4834°) - sin(65°)sin(0.4834°)≈0.42260.99996 - 0.90630.0084≈0.4226 - 0.0076≈0.415.

\(\sin(65.4834^\circ)\approx\sin(65.4834)\approx0.910\) (since sin(65°)≈0.9063, sin(65.4834°)≈sin(65°+0.4834°)=sin65°cos0.4834 + cos65°sin0.4834≈0.90630.99996 + 0.42260.0084≈0.9063 + 0.0036≈0.9099.

Now, c = 3445.3 m.

Then a = c cos B ≈ 3445.3 0.415 ≈ 3445.3 0.415 ≈ 1429.6 (since 3445 0.415 = 34450.4 + 34450.015 = 1378 + 51.675 = 1429.675), which matches option C's a=1,429.65.

Then b = c sin B ≈ 3445.3 0.9099 ≈ 3445.3 * 0.91 ≈ 3135.22, which is close to option C's b=3,134.68 (due to more precise calculation of sin(65.4834°)).

Also, angle A = 90° - 65.4834° = 24.5166°, which matches all options.

So the correct option is C.

Answer:

C. \( A = 24.5166^\circ, a = 1,429.65 \, \text{m}, b = 3,134.68 \, \text{m} \)