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CIE 9709 2022 March Paper 32 Q5

A Level / CIE / P3

CIE 9709 2022 March Paper 32 Paper · Question 5

题目

Problem

The angles α\alpha and β\beta lie between 0∘0^\circ and 180∘180^\circ and are such that

tan⁡(α+β)=2andtan⁡α=3tan⁡β.\tan(\alpha+\beta)=2\qquad\text{and}\qquad\tan\alpha=3\tan\beta.

Find the possible values of α\alpha and β\beta.

(6)
题目中文翻译

角 α\alpha 和 β\beta 均位于 0∘0^\circ 与 180∘180^\circ 之间，并满足

tan⁡(α+β)=2,tan⁡α=3tan⁡β.\tan(\alpha+\beta)=2,\qquad\tan\alpha=3\tan\beta.

求 α\alpha 和 β\beta 的可能值。

(6)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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令 t=tan⁡βt=\tan\beta，则 tan⁡α=3t\tan\alpha=3t。代入正切和角公式得到二次方程；每个 tt 对应一组角，不能把两组角任意配对。

答题过程

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Put t=tan⁡βt=\tan\beta, so tan⁡α=3t\tan\alpha=3t. The addition formula gives

3t+t1−3t2=2.\frac{3t+t}{1-3t^2}=2.

Hence

4t= 2−6t2,3t2+2t−1= 0,\begin{align*} 4t=&\,2-6t^2,\\[4mm] 3t^2+2t-1=&\,0, \end{align*}

so

(3t−1)(t+1)=0.(3t-1)(t+1)=0.

If t=1/3t=1/3, then

β= tan⁡−113=18.4349…∘,tan⁡α= 1,α=45∘.\begin{align*} \beta=&\,\tan^{-1}\frac13=18.4349\ldots^\circ,\\[4mm] \tan\alpha=&\,1,\quad\alpha=45^\circ. \end{align*}

If t=−1t=-1, then β=135∘\beta=135^\circ and tan⁡α=−3\tan\alpha=-3, giving

α=180∘−tan⁡−13=108.4349…∘.\alpha=180^\circ-\tan^{-1}3=108.4349\ldots^\circ.

Thus the possible pairs are

(α,β)≈(45∘,18.4∘)(\alpha,\beta)\approx(45^\circ,18.4^\circ)

or

(α,β)≈(108.4∘,135∘).(\alpha,\beta)\approx(108.4^\circ,135^\circ).

For both pairs, 1−tan⁡αtan⁡β≠01-\tan\alpha\tan\beta\ne0, so the sum tangent is defined.