Answer:
1) ∠B = 37°, a = 11
2) Area = 510 square units
Step-by-step explanation:
Question 1
Given angles and side lengths of triangle ABC:
- A = 30.5°
- C = 112.5°
- b = 13
Angles in a triangle sum to 180°. Therefore:
[tex]A+B+C=180^{\circ}\\\\30.5^{\circ}+B+112.5^{\circ}=180^{\circ}\\\\B+143^{\circ}=180^{\circ}\\\\B=37^{\circ}[/tex]
To find the length of side a, we can use the Law of Sines.
The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant, and this ratio is equal for all three sides:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
Substitute the values of A, B, and b into the Law of Sines equation and solve for side length a:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\\\\\\\dfrac{a}{\sin 30.5^{\circ}}=\dfrac{13}{\sin 37^{\circ}}\\\\\\a=\dfrac{13\sin 30.5^{\circ}}{\sin 37^{\circ}}\\\\\\a=10.96349952...\\\\\\a=11[/tex]
[tex]\hrulefill[/tex]
Question 2
Given angles and side lengths of triangle ABC:
- A = 30.5°
- C = 112.5°
- a = 30.5
To find the area of triangle ABC, we first need to find the length of side b.
Angles in a triangle sum to 180°. Therefore:
[tex]A+B+C=180^{\circ}\\\\30.5^{\circ}+B+112.5^{\circ}=180^{\circ}\\\\B+143^{\circ}=180^{\circ}\\\\B=37^{\circ}[/tex]
To find the length of side b, we can use the Law of Sines.
Substitute the values of A, B, and a into the Law of Sines equation and solve for side length b:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\\\\\\\dfrac{30.5}{\sin 30.5^{\circ}}=\dfrac{b}{\sin 37^{\circ}}\\\\\\b=\dfrac{30.5\sin 37^{\circ}}{\sin 30.5^{\circ}}\\\\\\b=36.1654596887...[/tex]
To find the area of a triangle given the measures of two of its sides and the enclosed angle, we can use the sine rule:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Area of a Triangle - Sine Rule}}\\\\\textsf{Area}=\dfrac{1}{2}ab \sin C\\\\\textsf{where:}\\ \phantom{ww}\bullet \;\textsf{$C$ is the angle.} \\ \phantom{ww}\bullet \;\textsf{$a$ and $b$ are the sides enclosing the angle.}\end{array}}[/tex]
In this case:
- a = 30.5
- b = 36.1654596887...
- C = 112.5°
Substitute the values into the formula and solve for area:
[tex]\textsf{Area}=\dfrac{1}{2}(30.5)(36.1654...) \sin 112.5^{\circ}\\\\\\\textsf{Area}=509.54105185...\\\\\\\textsf{Area}=510\; \sf square\;units[/tex]
Therefore, the area of triangle ABC is 510 square units, rounded to the nearest whole number.
Additional Notes
It is common practice to delay rounding until the final step to help maintain precision and reduce the risk of cumulative rounding errors. Therefore, I have used the exact value of b in the calculation for the area of the triangle. If your teacher wishes you to use the rounded value of b = 36, then the area would be 507 square units. However, please note that this is not considered best practice.
