Question

answer the following questions. make sure to explain all your work and show all your steps.
problem #1:
solve the following triangle for side a.

problem #2:
solve for angle a in the following triangle.

Explanation:

Problem #1

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle (SAS: side AC = 7, side AB = 9, angle at A = 35°). So we use the Law of Cosines. The Law of Cosines formula for side \( a \) (opposite angle A? Wait, no, in standard notation, side \( a \) is opposite angle \( A \), side \( b \) opposite angle \( B \), side \( c \) opposite angle \( C \). Wait, in the diagram, side \( a \) is BC, side AC is 7, side AB is 9, angle at A is 35°. So side \( a \) is BC, so by Law of Cosines: \( a^2 = b^2 + c^2 - 2bc \cos A \), where \( b = AC = 7 \), \( c = AB = 9 \), \( A = 35^\circ \). Wait, no, standard notation: in triangle ABC, side \( a \) is BC, side \( b \) is AC, side \( c \) is AB. So angle at A is between sides \( b \) and \( c \), so Law of Cosines: \( a^2 = b^2 + c^2 - 2bc \cos A \). Wait, no, Law of Cosines is \( a^2 = b^2 + c^2 - 2bc \cos A \) where \( a \) is opposite angle \( A \)? No, wait, no: Law of Cosines is \( c^2 = a^2 + b^2 - 2ab \cos C \), where \( C \) is the angle between sides \( a \) and \( b \). So in this case, angle at A is 35°, between sides AC (length 7) and AB (length 9). So the side opposite angle A is BC, which is \( a \). Wait, no, angle at A: sides adjacent are AC (7) and AB (9), side opposite is BC (a). So Law of Cosines: \( a^2 = AC^2 + AB^2 - 2 \times AC \times AB \times \cos(\angle A) \). So \( a^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(35^\circ) \).

Step2: Calculate Each Term

First, calculate \( 7^2 = 49 \), \( 9^2 = 81 \), so \( 49 + 81 = 130 \). Then, \( 2 \times 7 \times 9 = 126 \). \( \cos(35^\circ) \approx 0.8192 \). So \( 126 \times 0.8192 \approx 126 \times 0.8192 \approx 103.2192 \). Then, \( a^2 = 130 - 103.2192 = 26.7808 \)? Wait, that can't be right, because 7 and 9 with included angle 35°, the side opposite should be longer? Wait, no, I think I mixed up the sides. Wait, in the diagram, side AC is 7, side BC is \( a \), side AB is 9, angle at A is 35°. Wait, maybe I got the sides wrong. Wait, the triangle: A---B is 9, A---C is 7, angle at A is 35°, so B---C is \( a \). So yes, angle at A is between AB (9) and AC (7), so side BC (a) is opposite angle A? No, angle at A is between AB and AC, so side BC is opposite angle A? Wait, no, angle at A: vertices A, B, C. So angle at A is between AB and AC, so side BC is opposite angle A. So Law of Cosines: \( BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle A) \). So \( a^2 = 9^2 + 7^2 - 2 \times 9 \times 7 \times \cos(35^\circ) \). Let's recalculate:

\( 9^2 = 81 \), \( 7^2 = 49 \), so \( 81 + 49 = 130 \). \( 2 \times 9 \times 7 = 126 \). \( \cos(35^\circ) \approx 0.819152 \). So \( 126 \times 0.819152 \approx 126 \times 0.819152 \approx 103.213152 \). Then, \( a^2 = 130 - 103.213152 = 26.786848 \). Then, \( a = \sqrt{26.786848} \approx 5.175 \). Wait, that seems too short. Wait, maybe I mixed up the sides. Wait, maybe the side labeled 7 is BC, and side a is AC? No, the diagram shows: A to B is 9, A to C is 7, B to C is a, angle at A is 35°. So maybe I made a mistake in the Law of Cosines. Wait, no, Law of Cosines: if we have two sides and the included angle, then the third side is calculated by \( c^2 = a^2 + b^2 - 2ab \cos C \), where \( C \) is the included angle. So in this case, included angle is at A, between sides AB (9) and AC (7), so side BC (a) is opposite? No, included angle is between AB and AC, so the side opposite the included angle is BC? No, included angle is between AB and AC, so the side opposite is BC? Wait, no, included angle is between two sides, so the third side is opposite t…

Step1: Identify the Law to Use

We have a triangle with all three sides: AB = 10, AC = 7, BC = 8. So we use the Law of Cosines to find angle A. The Law of Cosines formula for angle A is \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \), where \( a = BC = 8 \), \( b = AC = 7 \), \( c = AB = 10 \).

Step2: Substitute the Values

Substitute \( a = 8 \), \( b = 7 \), \( c = 10 \) into the formula:

\( \cos A = \frac{7^2 + 10^2 - 8^2}{2 \times 7 \times 10} \)

Step3: Calculate the Numerator and Denominator

First, calculate the numerator:

\( 7^2 = 49 \), \( 10^2 = 100 \), \( 8^2 = 64 \)

\( 49 + 100 - 64 = 149 - 64 = 85 \)

Denominator:

\( 2 \times 7 \times 10 = 140 \)

So \( \cos A = \frac{85}{140} \approx 0.6071 \)

Step4: Find Angle A

Take the arccosine of 0.6071:

\( A = \arccos(0.6071) \approx 52.6^\circ \) (rounded to one decimal place)

Answer:

\( a \approx 5.18 \) (rounded to two decimal places)

Problem #2