If the length of side IJ is 2x – 2, the length of side IK is 2x + 1, the length of side KM is 3x – 1, and the length of side LM is 2x + 2, what is the length of side LM?
Line EF is parallel to line GH, and line JL is perpendicular to lines EF and GH. The diagram is not to scale.
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Solution
Angles IKJ and LKM are vertical angles; their measures are equal. Since the sine of an angle is equal to the length of the side opposite the angle divided by the length of the hypotenuse, the sine of angle IKJ is equal to \(\frac{\overline{IJ}}{\overline{IK}}\) and the sine of angle LKM is equal to\(\frac{\overline{LM}}{\overline{KM}}\).Two equal angles have equal sines;\(\frac{\overline{IJ}}{\overline{IK}}=\frac{\overline{LM}}{\overline{KM}},\frac{2x-2}{2x+1}=\frac{2x+2}{3x+1}\), 6x2 – 6x – 2x + 2 = 4x2 + 2x + 4x + 2, 6x2 – 8x + 2 = 4x2 + 6x + 2, 2x2– 14x = 0, x2 – 7x = 0, x(x – 7) = 0, x = 7 (x cannot be 0 since the length of a line cannot be negative—if x were 0, \(\overline{IJ}\) would be –2 units). If x = 7, then the length of \(\overline{LM}\)= 2(7) + 2 = 14 + 2 = 16 units.
If angle KMH is three times the size of angle KML and the length of side JK is x√6 units, what is the length of side IK?
Line EF is parallel to line GH, and line JL is perpendicular to lines EF and GH. The diagram is not to scale.
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Solution
Angles KMH and KML are supplementary angles; their measures add to 180°. If the measure of angle KML is y and the measure of angle KMH is 3y, then 4y = 180, y = 45°. The measure of angle KML is 45°. Angle JIK is also 45°, since KML and JIK are alternating angles. Since JIK is 45° and KJI is 90°, angle IKJ is 45° (180 – (45 + 90) = 45). Therefore, triangle IJK is a 45- 45-90 right triangle. The hypotenuse of a 45- 45-90 right triangle is √2 times the length of either base: (x√6)(√2) = x12 = x√4√3 = 2x√3 units.
If the tangent of angle JIK is √3, then
Line EF is parallel to line GH, and line JL is perpendicular to lines EF and GH. The diagram is not to scale.
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Solution
The tangent of a 60° angle is √3; therefore, the measure of angle JIK is 60°. Angles JIK and LMK are alternating angles; their measures are equal. Angle LMK is 60° and its tangent is √3. Therefore, the measurec of \(\overline{LK}\) , the side opposite angle LMK, is √3 times the length of \(\overline{LM}\), the side adjacent to angle LMK.
If the length of side LM is 8 units and the sum of angles IKJ and LKM is 60°, what is the length of side KM?
Line EF is parallel to line GH, and line JL is perpendicular to lines EF and GH. The diagram is not to scale.
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Solution
Angles IKJ and LKM are vertical angles; their measures are equal. Therefore, each angle is equal to \(\frac{60}{2}\) = 30°. The sine of a 30° angle is 1⁄2. Since the measure of angle LKM is 30°, the length of \(\overline{LM}\) divided by the length of \(\overline{KM}\) is equal to 1⁄2: \(\frac{\overline{LM}}{\overline{KM}}\) = 1⁄2,\(\frac{8}{\overline{KM}}\) = 1⁄2,\(\overline{KM}\)= 16 units.
If the length of side KM is 10 units and the length of side LM is 5 units, what is the measure of angle KIJ?
Line EF is parallel to line GH, and line JL is perpendicular to lines EF and GH. The diagram is not to scale.
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Solution
Since \(\overline{KM}\)= 10 and \(\overline{KM}\)= 5, the cosine of angle LMK is = 1⁄2. The cosine of a 60° angle is 1⁄2, so the measure of angle LMK is 60°. Angles LMK and KIJ are alternating angles, so their measures are equal. Therefore, angle KIJ is also 60°.