Question

Shown is a circle, centre C, with diameter AE.

PQ is a tangent to the circle, meeting it at E.
A EDQ is isosceles with ED= EQ.
Given that / EAD = 50°, calculate the size
of Z DQE.

Answer 1

∠DQE = 65°

Explanation:

You want to know the measure of Angle DQE in the given figure.

Right triangle

Triangle ADE is inscribed in a semicircle, so is a right triangle. That makes acute angle AED complementary to acute angle EAD. The latter is given as 50°, so ...

∠AED = 90° -50° = 40°

Tangent

We know that tangent line QE is perpendicular to diameter AE, so angle AEQ is a right angle. Segment ED divides that right angle into complementary angles, so ...

∠DEA +∠DEQ = 90°

∠DEQ = 90° -∠DEA = 90° -40° = 50°

Isosceles triangle

A.pex angle DEQ in isosceles triangle DEQ being 50° means that the base angles QDE and DQE will have measure ...

∠DQE = (180° -50°)/2 = 130°/2

∠DQE = 65°

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Alternate solution

Let angles DQE and QDE be represented by x. Then the sum of angles in quadrilateral ADQE is 360°:

50° +(90+x)° +x° +90° = 360°

230° +2x = 360° . . . . . . simplify

x = 130°/2 = 65° . . . . . solve for x