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Pre-practice (Optional):Find the missing side length in each triangle, to the nearest tenth of a unit.342900062865009048755842000 b) Solve for the indicated angle, to the nearest degree.790575990600035814004191000 b) Mandatory Practice (no calculator):Sketch each triangle and use the given information to find the missing side length, to the nearest tenth of a unit.In acute ?TUV, t=1.8 cm, v=1.4 cm, and U=52°.In acute ?DEF, e=1.1 km, f=1.6 km, and D=54°.Solve each triangle.381000046990007905759461500 b) A distress signal is received from a ship that is 21 km from one port and 17 km from another port. The eastern port is 24 km directly east of the western port. At what angle to the western shoreline should the ship head, in order to dock at the western port? Round to the nearest degree.For the triangle A(4,0), B(0,4) and C(0,0), find the equation for the median from vertex C.For the triangle A(4,0), B(0,4) and C(0,0), find the equation for the right bisector through AB. Explain any similarity to your answer from the previous question. Factor 3x(x+6)-6(x+6)Solve the linear system 3x+4y=27 and x-2y=-1.Evaluate 34×1-115÷15Challenge Acute ?ABC is isosceles, with b=c. Show that cosA=1-a22b2.Solve ?ABC, if the equal sides are 1.5 cm and the third side is 0.8 cm.Draw an equilateral triangle and label its sides a. Mark one of the angles 60°. Use the cosine law to prove that cos60°=12Answers:a) 4.3 m b) 1.7 cma) 51° b) 66° a) u=1.4 cm b) d=1.7 cm a) r=16 cm, P=48° , Q=62° b) V=79°, T=57°, U=44°31° y=xy=x. The median from the previous questions actually bisects the side AB at a right angle. Thus the median and right bisector are the same line in this situation. This is because the triangle ABC is an isosceles. 3(x+6)(x-2)x=5, y=3 72b) A=30.9°, B=C=74.55° ................
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