解三角形六題

1. △ABC中，∠A=105°，∠B=60°，c=4，求∠C，a，b。

Sol.
∠C = 180°-105°-60° = 15° (∠sum of △)
a/sin ∠A = c/sin ∠C (sine law)
a/sin 105° = 4/sin 15°
a = 14.9, corr. to 3 sig. fig.

b/sin ∠B = c/sin ∠C (sine law)
b/sin 60° = 4/sin 15°
b = 13.4, corr. to 3 sig. fig.


2. △ABC中，a=(√3)+1，b=√2，c=2，求∠A，∠B，∠C。

Sol.
a^2 = b^2 + c^2 - 2bc cos ∠A (cosine law)
[(√3)+1]^2 = (√2)^2 + 2^2 - 2(√2)(2) cos ∠A
∠A = 105°

b^2 = a^2 + c^2 - 2ac cos ∠B (cosine law)
(√2)^2 = [(√3)+1]^2 + 2^2 - 2[(√3)+1](2) cos ∠B
∠B = 30°

∠C = 180°-105°-30° = 45° (∠sum of △)


3. △ABC中，a=x^2+x+1，b=x^2-1，c=2x+1，求其最大角之度數。

Sol.
From b, x must be greater than 1.
Thus, a is the longest.
Thus, ∠A is largest. (greater side, greater ∠)
cos ∠A = (a^2 - b^2 - c^2) / (-2bc) (cosine law)
= [(x^2+x+1)^2 - (x^2-1)^2 - (2x+1)^2] / -2(x^2-1)(2x+1)
= -1/2
∠A = 120°


4. △ABC中，b=5，c=3，∠A=120°，求a，∠B，∠C。

Sol.
a^2 = b^2 + c^2 - 2bc cos ∠A (cosine law)
a^2 = 5^2 + 3^2 - 2(5)(3) cos 120°
a = 7

a/sin ∠A = b/sin ∠B (sine law)
7/sin 120° = 5/sin ∠B
∠B = 38.2°, corr. to 3 sig. fig.

a/sin ∠A = c/sin ∠C (sine law)
7/sin 120° = 3/sin ∠C
∠C = 21.8°, corr. to 3 sig. fig.


5. △ABC中，b=(2√3)，c=(3√2)，∠C=60°，求a，∠A，∠B。

Sol.
c/sin ∠C = b/sin ∠B (sine law)
3√2/sin 60° = 2√3/sin ∠B
∠B = 45°

∠A = 180° - 60° - 45° = 75°

c/sin ∠C = a/sin ∠A (sine law)
3√2/sin 60° = a/sin 75°
a = 4.73, corr. to 3 sig. fig.


6. △ABC中，∠A=15°，a=3-(√3)，b=3+(√3)，求c，∠B，∠C。

Sol.
a/sin ∠A = b/sin ∠B (sine law)
[3-(√3)] /sin15° = [3+(√3)] /sin ∠B
∠B = 75°

∠C = 180° - 15° - 75° = 90°

c^2 = a^2 + b^2 (Pyth. theorem)
= [3-(√3)]^2 + [3+(√3)]^2
c = 4.90, corr. to 3 sig. fig.