Let ABC be an isosceles triangle with AB = AC and ∠BAC=30°. Side AB and the median AD contain points Q and P respectively (point
s P and Q are not at point A) in such a way that PC=PQ. Find m∠PQC.
1 answer:
Answer:
15°
Step-by-step explanation:
Since P is on the median of ΔABC, it is equidistant from points B and C as well as from C and Q. Thus, points B, C, and Q all lie on a circle centered at P. (See the attached diagram.)
The base angles (B and C) of triangle ABC are (180° -30°)/2 = 75°. This means arc QC of the circle centered at P has measure 150°. The diameter of circle P that includes point Q is defined to intersect circle P at R.
Central angle RPC is the difference between arcs QR and QC, so is 180° -150° = 30°. Inscribed angle RQC has half that measure, so is 15°. Angle PQC has the same measure as angle RQC, so is 15°.
Angle PQC is 15°.
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Answer: < -4/5, 3/5>
This is equivalent to writing < -0.8, 0.6 >
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Explanation:
Draw an xy grid and plot the point (-4,3) on it. Draw a segment from the origin to this point. Then draw a vertical line until reaching the x axis. See the diagram below.
We have a right triangle with legs of 4 and 3. The hypotenuse is through use of the pythagorean theorem.
We have a 3-4-5 right triangle.
Therefore, the vector is 5 units long. This is the magnitude of the vector.
Divide each component by the magnitude so that the resulting vector is a unit vector pointing in this same direction.
Therefore, we go from < -4, 3 > to < -4/5, 3/5 >
This is equivalent to < -0.8, 0.6 > since -4/5 = -0.8 and 3/5 = 0.6
Side note: Unit vectors are useful in computer graphics.
