Question

6
multiple choice 1 point
two triangles can be formed using the given measurements. solve both triangles.
b = 40°, b = 26, c = 29
a = 95.8°, c = 44.2°, a = 40.2; a = 84.2°, c = 135.8°, a = 40.2
a = 94.2°, c = 45.8°, a = 40.3; a = 5.8°, c = 134.2°, a = 4.1
a = 95.8°, c = 44.2°, a = 16.8; a = 84.2°, c = 135.8°, a = 16.8
a = 94.2°, c = 45.8°, a = 16.8; a = 5.8°, c = 134.2°, a = 16.8
7
multiple choice 1 point
two triangles can be formed using the given measurements. solve both triangles.
c = 72°, a = 28, c = 27
a = 9.5°, b = 98.5°, b = 26; a = 170.5°, b = 81.5°, b = 26
a = 80.5°, b = 27.5°, b = 55.6; a = 99.5°, b = 8.5°, b = 55.6
a = 9.5°, b = 98.5°, b = 28.1; a = 170.5°, b = 81.5°, b = 28.1
a = 80.5°, b = 27.5°, b = 13.1; a = 99.5°, b = 8.5°, b = 4.2

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). For the first triangle (Problem 6):
Given \(B = 40^{\circ}\), \(b = 26\), \(c = 29\). First, find \(\sin C\) using \(\frac{b}{\sin B}=\frac{c}{\sin C}\), so \(\sin C=\frac{c\sin B}{b}\).
Substitute the values: \(\sin C=\frac{29\sin40^{\circ}}{26}\approx\frac{29\times0.6428}{26}\approx0.719\). Then \(C_1=\sin^{- 1}(0.719)\approx46^{\circ}\) (approximate value for calculation, actual value adjusted for options).
Using \(A + B + C=180^{\circ}\), \(A_1 = 180^{\circ}-40^{\circ}-C_1\). And \(a=\frac{b\sin A}{\sin B}\).
For the second - triangle case (ambiguous case of the Law of Sines), \(C_2 = 180^{\circ}-C_1\). Then \(A_2=180^{\circ}-B - C_2\) and \(a=\frac{b\sin A}{\sin B}\).
After detailed calculation (using \(\sin40^{\circ}\approx0.6428\), \(\sin45.8^{\circ}\approx0.717\), \(\sin94.2^{\circ}\approx0.997\), \(\sin5.8^{\circ}\approx0.101\)):
If \(B = 40^{\circ}\), \(b = 26\), \(c = 29\)
\(\sin C=\frac{c\sin B}{b}=\frac{29\times\sin40^{\circ}}{26}\). \(\sin40^{\circ}\approx0.6428\), \(\frac{29\times0.6428}{26}\approx0.717\), \(C\approx45.8^{\circ}\) or \(C = 180 - 45.8=134.2^{\circ}\)
When \(C = 45.8^{\circ}\), \(A=180-(40 + 45.8)=94.2^{\circ}\), \(a=\frac{b\sin A}{\sin B}=\frac{26\times\sin94.2^{\circ}}{\sin40^{\circ}}\approx\frac{26\times0.997}{0.6428}\approx40.3\)
When \(C = 134.2^{\circ}\), \(A=180-(40 + 134.2)=5.8^{\circ}\), \(a=\frac{b\sin A}{\sin B}=\frac{26\times\sin5.8^{\circ}}{\sin40^{\circ}}\approx\frac{26\times0.101}{0.6428}\approx4.1\)

Step2: Check the second problem (Problem 7)

Given \(C = 72^{\circ}\), \(a = 28\), \(c = 27\)
Using the Law of Sines \(\frac{a}{\sin A}=\frac{c}{\sin C}\), \(\sin A=\frac{a\sin C}{c}=\frac{28\sin72^{\circ}}{27}\)
\(\sin72^{\circ}\approx0.9511\), \(\sin A=\frac{28\times0.9511}{27}\approx0.98\)
\(A_1=\sin^{-1}(0.98)\approx80.5^{\circ}\), then \(B_1=180-(72 + 80.5)=27.5^{\circ}\), \(b=\frac{c\sin B}{\sin C}=\frac{27\sin27.5^{\circ}}{\sin72^{\circ}}\approx\frac{27\times0.462}{0.9511}\approx13.1\)
For the second - triangle case (ambiguous case), \(A_2 = 180 - 80.5=99.5^{\circ}\), \(B_2=180-(72 + 99.5)=8.5^{\circ}\), \(b=\frac{c\sin B}{\sin C}=\frac{27\sin8.5^{\circ}}{\sin72^{\circ}}\approx\frac{27\times0.148}{0.9511}\approx4.2\)

Answer:

For Problem 6: \(A = 94.2^{\circ},C = 45.8^{\circ},a = 40.3;A = 5.8^{\circ},C = 134.2^{\circ},a = 4.1\)
For Problem 7: \(A = 80.5^{\circ},B = 27.5^{\circ},b = 13.1;A = 99.5^{\circ},B = 8.5^{\circ},b = 4.2\)