Question

solve the triangle.
b = 72°15, c = 24 m, a = 70 m
what is the length of side b?
b = □ m
(round to the nearest whole number as needed.)
what is the measure of angle a?
a = □°□
(round to the nearest whole number as needed.)
what is the measure of angle c?
c = □°□
(round to the nearest whole number as needed.)

Explanation:

Part 1: Find the length of side \( b \)

We use the Law of Cosines, which states that for a triangle with sides \( a \), \( b \), \( c \) and the angle \( B \) opposite side \( b \), the formula is:

$$ b^2 = a^2 + c^2 - 2ac \cos(B) $$

First, convert \( B = 72^\circ 15' \) to decimal degrees. Since \( 15' = \frac{15}{60} = 0.25^\circ \), so \( B = 72.25^\circ \).

Now, substitute \( a = 70 \), \( c = 24 \), and \( B = 72.25^\circ \) into the formula:

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Calculate \( \cos(72.25^\circ) \approx 0.3057 \):

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Take the square root of \( 4448.848 \) to find \( b \):

$$ b = \sqrt{4448.848} \approx 66.7 \approx 67 $$

Part 2: Find the measure of angle \( A \)

We use the Law of Sines, which states that \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \). So,

$$ \sin(A) = \frac{a \sin(B)}{b} $$

We know \( a = 70 \), \( b \approx 67 \), and \( B = 72.25^\circ \).

$$ \sin(A) = \frac{70 \times \sin(72.25^\circ)}{67} $$

Calculate \( \sin(72.25^\circ) \approx 0.9526 \):

$$ \sin(A) = \frac{70 \times 0.9526}{67} = \frac{66.682}{67} \approx 0.9953 $$

Then, \( A = \arcsin(0.9953) \approx 84.5^\circ \). Convert \( 0.5^\circ \) to minutes: \( 0.5 \times 60 = 30' \), so \( A \approx 84^\circ 30' \).

Part 3: Find the measure of angle \( C \)

We know that the sum of the angles in a triangle is \( 180^\circ \). So,

$$ C = 180^\circ - A - B $$

Substitute \( A \approx 84.5^\circ \) and \( B = 72.25^\circ \):

$$ C = 180^\circ - 84.5^\circ - 72.25^\circ = 23.25^\circ $$

Convert \( 0.25^\circ \) to minutes: \( 0.25 \times 60 = 15' \), so \( C = 23^\circ 15' \).

Final Answers

• Length of side \( b \): \( \boxed{67} \) m
• Measure of angle \( A \): \( \boxed{84^\circ 30'} \)
• Measure of angle \( C \): \( \boxed{23^\circ 15'} \)

Answer:

Part 1: Find the length of side \( b \)

We use the Law of Cosines, which states that for a triangle with sides \( a \), \( b \), \( c \) and the angle \( B \) opposite side \( b \), the formula is:

$$ b^2 = a^2 + c^2 - 2ac \cos(B) $$

First, convert \( B = 72^\circ 15' \) to decimal degrees. Since \( 15' = \frac{15}{60} = 0.25^\circ \), so \( B = 72.25^\circ \).

Now, substitute \( a = 70 \), \( c = 24 \), and \( B = 72.25^\circ \) into the formula:

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Calculate \( \cos(72.25^\circ) \approx 0.3057 \):

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Take the square root of \( 4448.848 \) to find \( b \):

$$ b = \sqrt{4448.848} \approx 66.7 \approx 67 $$

Part 2: Find the measure of angle \( A \)

We use the Law of Sines, which states that \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \). So,

$$ \sin(A) = \frac{a \sin(B)}{b} $$

We know \( a = 70 \), \( b \approx 67 \), and \( B = 72.25^\circ \).

$$ \sin(A) = \frac{70 \times \sin(72.25^\circ)}{67} $$

Calculate \( \sin(72.25^\circ) \approx 0.9526 \):

$$ \sin(A) = \frac{70 \times 0.9526}{67} = \frac{66.682}{67} \approx 0.9953 $$

Then, \( A = \arcsin(0.9953) \approx 84.5^\circ \). Convert \( 0.5^\circ \) to minutes: \( 0.5 \times 60 = 30' \), so \( A \approx 84^\circ 30' \).

Part 3: Find the measure of angle \( C \)

We know that the sum of the angles in a triangle is \( 180^\circ \). So,

$$ C = 180^\circ - A - B $$

Substitute \( A \approx 84.5^\circ \) and \( B = 72.25^\circ \):

$$ C = 180^\circ - 84.5^\circ - 72.25^\circ = 23.25^\circ $$

Convert \( 0.25^\circ \) to minutes: \( 0.25 \times 60 = 15' \), so \( C = 23^\circ 15' \).

Final Answers

• Length of side \( b \): \( \boxed{67} \) m
• Measure of angle \( A \): \( \boxed{84^\circ 30'} \)
• Measure of angle \( C \): \( \boxed{23^\circ 15'} \)