Using the Law of Cosines in a Multiple Triangle Figure

Question

In the figure, triangle ABC is a right triangle, CQ = 4, and BQ = 8. Also, ∠AQC = 30º and ∠CQB = 45º. Find the length of AQ. [Hint: first use the Law of Cosines to find expressions for a², b², and c².]

Essentially what the question is asking is solve, algebraically for x, but I am completely stumped on how to do so.

Any help would be greatly appreciated.

Answer 1

Consider the triangle AQBC.

By the law of cosines, we get BC²=QC²+QB²-2(QC)(QB) cos ZBOC a² = 8² +4²-2(8) (4) cos 45º

a²-80-64 = (1/√2)

a²=80-32√2

Consider the triangle AQAC.

By the law of cosines, we get

AC² = QC²+Q4²-2(QC) (04) cos ZAOC

b² = 8² + x²-2(8)(x) cos 30°

b² = 64+x²-16x(√3/2)

b² = 64+x²-8√3x

Consider the triangle AQBA.

By the law of cosines, we get

AB²=QB²+QA²-2(OB)(QA) cos LAQB

c²=4²+x²-2 (4)(x) cos(30+45)°

c²16+x²-8x (0.2588)

c² = 16+x² - 2.0704x

Consider the triangle AABC.

By the Pythagoras theorem, we get

AB²= BC² + AC²

c² = a² + b²

16+ x² -2.0704x = 80-32√2+64 + x² -8√3x

8√3x-2.0704x=128-32√2

x(8√3-2.0704)=128-32-√2

x(128-32√2)/(8√3-2.0704)

x ≈7.0206

Hence, AQ approximately equal to 7.0206.