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Solving Triangles

12 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0532 International

Problem 1 — Solving Triangles

In △ABC\triangle ABC, AB=6AB = 6, AC=8AC = 8, and ∠A=π6\angle A = \dfrac{\pi}{6}. The area equals:

Show answer & worked solution
  1. A. 4848
  2. B. 2424
  3. C. 1212✓ correct
  4. D. 66

Area =12⋅6⋅8⋅sin⁡π6=24⋅12=12= \dfrac{1}{2} \cdot 6 \cdot 8 \cdot \sin\dfrac{\pi}{6} = 24 \cdot \dfrac{1}{2} = 12.

Problem #0533 International

Problem 2 — Solving Triangles

In △ABC\triangle ABC, a=7a = 7, b=8b = 8, c=5c = 5. The value of cos⁡C\cos C is:

Show answer & worked solution
  1. A. 114\dfrac{1}{14}
  2. B. 1116\dfrac{11}{16}
  3. C. 1114\dfrac{11}{14}✓ correct
  4. D. 58\dfrac{5}{8}

25=49+64−112cos⁡C⇒112cos⁡C=88⇒cos⁡C=88112=111425 = 49 + 64 - 112\cos C \Rightarrow 112\cos C = 88 \Rightarrow \cos C = \dfrac{88}{112} = \dfrac{11}{14}.

Problem #0531 International

Problem 3 — Solving Triangles

A right triangle has legs 33 and 44. Its area equals:

Show answer & worked solution
  1. A. 55
  2. B. 66✓ correct
  3. C. 77
  4. D. 1212

Area =12⋅3⋅4=6= \dfrac{1}{2} \cdot 3 \cdot 4 = 6.

Problem #0535 International

Problem 4 — Solving Triangles

In △ABC\triangle ABC, AB=5AB = 5, BC=12BC = 12, AC=13AC = 13. The area equals:

Show answer & worked solution
  1. A. 2020
  2. B. 2424
  3. C. 3030✓ correct
  4. D. 6060

Since 52+122=1325^2 + 12^2 = 13^2, △ABC\triangle ABC is right-angled at BB. Area =12⋅5⋅12=30= \dfrac{1}{2} \cdot 5 \cdot 12 = 30.

Problem #0537 International

Problem 5 — Solving Triangles

In △ABC\triangle ABC, a=6a = 6 and ∠A=π6\angle A = \dfrac{\pi}{6}. The circumradius RR equals:

Show answer & worked solution
  1. A. 33
  2. B. 3\sqrt{3}
  3. C. 66✓ correct
  4. D. 1212

2R=asin⁡A=61/2=12⇒R=62R = \dfrac{a}{\sin A} = \dfrac{6}{1/2} = 12 \Rightarrow R = 6.

Problem #0536 International

Problem 6 — Solving Triangles

The area of a triangle with sides 55, 66, 77 is closest to:

Show answer & worked solution
  1. A. 1010
  2. B. 1212
  3. C. 14.714.7✓ correct
  4. D. 2121

s=5+6+72=9s = \dfrac{5 + 6 + 7}{2} = 9. A=9⋅4⋅3⋅2=216=66≈14.70A = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt{6} \approx 14.70.

Problem #0538 International

Problem 7 — Solving Triangles

In △ABC\triangle ABC, a=5a = 5, b=6b = 6, c=7c = 7. The inradius rr is closest to:

Show answer & worked solution
  1. A. 1.41.4
  2. B. 1.61.6✓ correct
  3. C. 2.42.4
  4. D. 3.03.0

s=9s = 9. By Heron, Area =9⋅4⋅3⋅2=66≈14.70= \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = 6\sqrt{6} \approx 14.70. So r≈14.709≈1.63r \approx \dfrac{14.70}{9} \approx 1.63.

Problem #0534 International

Problem 8 — Solving Triangles

In △ABC\triangle ABC, ∠A=π6\angle A = \dfrac{\pi}{6}, ∠B=π4\angle B = \dfrac{\pi}{4}, and a=4a = 4. The side bb equals:

Show answer & worked solution
  1. A. 222\sqrt{2}
  2. B. 44
  3. C. 424\sqrt{2}✓ correct
  4. D. 828\sqrt{2}

4sin⁡(π/6)=bsin⁡(π/4)⇒41/2=b2/2⇒8=b2/2⇒b=8⋅22=42\dfrac{4}{\sin(\pi/6)} = \dfrac{b}{\sin(\pi/4)} \Rightarrow \dfrac{4}{1/2} = \dfrac{b}{\sqrt{2}/2} \Rightarrow 8 = \dfrac{b}{\sqrt{2}/2} \Rightarrow b = 8 \cdot \dfrac{\sqrt{2}}{2} = 4\sqrt{2}.

Problem #0939 US SAT

Problem 9 — Law of Cosines

In a triangle, two sides have lengths 44 and 55 and the included angle is 60∘60^{\circ}. The length of the third side is:

Show answer & worked solution
  1. A. 41\sqrt{41}
  2. B. 21\sqrt{21}✓ correct
  3. C. 61\sqrt{61}
  4. D. 33

c2=42+52−2(4)(5)cos⁡60∘=16+25−40⋅12=41−20=21c^{2} = 4^{2} + 5^{2} - 2(4)(5)\cos 60^{\circ} = 16 + 25 - 40 \cdot \tfrac{1}{2} = 41 - 20 = 21.

So c=21c = \sqrt{21}.

Problem #0938 US SAT

Problem 10 — Law of Sines

In a triangle, aa is opposite ∠A=30∘\angle A = 30^{\circ} and bb is opposite ∠B=90∘\angle B = 90^{\circ}. If a=5a = 5, then bb equals:

Show answer & worked solution
  1. A. 55
  2. B. 52\dfrac{5}{2}
  3. C. 1010✓ correct
  4. D. 535\sqrt{3}

5sin⁡30∘=bsin⁡90∘\dfrac{5}{\sin 30^{\circ}} = \dfrac{b}{\sin 90^{\circ}}.

sin⁡30∘=12\sin 30^{\circ} = \tfrac{1}{2} and sin⁡90∘=1\sin 90^{\circ} = 1, so 51/2=b1\dfrac{5}{1/2} = \dfrac{b}{1}.

b=10b = 10.

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