Question

directions: solve for x. then find all missing angles. triangles may not be drawn to scale.
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1. triangle abc has the following interior angle measures ( mangle abc = 20^{circ}, mangle bca = 64^{circ} ). find the measure of ( angle cab ).
2. triangle def has the following interior angle measures: ( mangle def=(4x)^{circ}, mangle efd = 22^{circ} ). if ( mangle fde = 118^{circ} ), solve for x and find all angles.

Explanation:

Step1: Use triangle sum theorem

The sum of interior angles of a triangle is \(180^{\circ}\). For problem 1: \(x + 50+58 = 180\)

Step2: Solve for \(x\)

\(x=180-(50 + 58)=180 - 108=72\)

For problem 2: It is an isosceles triangle (two sides are equal, so two base angles are equal). Let \(x\) be one of the base angles. Then \(x+x + 56=180\), \(2x=180 - 56=124\), \(x = 62\)

For problem 3: It is a right - triangle (\(90^{\circ}\) angle). So \(x+41 + 90=180\), \(x=180-(41 + 90)=49\)

For problem 4: \(3x+29+118 = 180\), \(3x=180-(29 + 118)=33\), \(x = 11\), then \(3x=33\)

For problem 5: Use the exterior - angle property. The exterior angle \((x + 35)\) is equal to the sum of the two non - adjacent interior angles. So \(x + 35=73+(x + 28)\) (This is wrong, we should use triangle sum. Let the third interior angle be \(y\), \(y=180-(73+(x + 28))\), and also \(x + 35=180 - y\). Substitute \(y\): \(x + 35=73+(x + 28)\) (error in approach, correct: sum of interior angles of triangle: \(73+(x + 28)+(180-(x + 35))=180\) (not helpful). Correct way: \(x+35\) (exterior angle) and sum of interior angles: Let the third angle be \(A\), \(A = 180-(73+(x + 28))\), and \(x + 35=180 - A\). \(x+35=73+(x + 28)\) (wrong). Correct: Using sum of interior angles of triangle: \(73+(x + 28)+(180-(x + 35))=180\) (not useful). Correct formula: \(x+35\) (exterior) and for the triangle, \(73+(x + 28)+y = 180\), \(y=180-(73+(x + 28))\), and \(x + 35=180 - y\). Substitute \(y\): \(x + 35=73+(x + 28)\) (no, correct: \(x+35\) (exterior) and sum of interior angles: Let the triangle have angles \(73\), \(x + 28\), \(z\). \(z=180-(73+(x + 28))\). Also, \(x + 35=z\) (exterior angle property). So \(x + 35=180-(73+(x + 28))\), \(x+35=180 - 73 - x - 28\), \(x+x=180 - 73 - 28 - 35\), \(2x=44\), \(x = 22\)

For problem 6: \(x+(x + 102)+(x + 27)=180\), \(3x+129 = 180\), \(3x=51\), \(x = 17\), then \(x + 102=119\), \(x + 27=44\)

For problem 7: \(m\angle CAB=180-(m\angle ABC + m\angle BCA)=180-(20 + 64)=96\)

For problem 8: \(4x+22 + 118=180\), \(4x=180-(22 + 118)=40\), \(x = 10\), \(4x = 40\)

Answer:

1. \(x = 72^{\circ}\), angles: \(50^{\circ}\), \(58^{\circ}\), \(72^{\circ}\)
2. \(x = 62^{\circ}\), angles: \(56^{\circ}\), \(62^{\circ}\), \(62^{\circ}\)
3. \(x = 49^{\circ}\), angles: \(41^{\circ}\), \(49^{\circ}\), \(90^{\circ}\)
4. \(x = 11\), angles: \(29^{\circ}\), \(33^{\circ}\), \(118^{\circ}\)
5. \(x = 22\)
6. \(x = 17\), angles: \(17^{\circ}\), \(44^{\circ}\), \(119^{\circ}\)
7. \(m\angle CAB = 96^{\circ}\)
8. \(x = 10\), angles: \(22^{\circ}\), \(40^{\circ}\), \(118^{\circ}\)