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Trigonometric Ratios of Some Particular Angles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Ratios of Some Particular Angles4 Marks
Question
Evaluate the following:
If $\tan(\text{A}-\text{B})=\frac{1}{\sqrt{3}}$ and $\tan(\text{A}+\text{B})=\sqrt{3},0^\circ<(\text{A}+\text{B})\leq90^\circ$ and $\text{A}>\text{B}$ then find A and B.
✓

Answer

Here, $\tan(\text{A}-\text{B})=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan(\text{A}-\text{B})=\tan30^\circ$ $\Big[\because\ \tan30^\circ=\frac{1}{\sqrt{3}}\Big]$
$\Rightarrow\text{A}-\text{B}=30^\circ\dots(\text{i})$
Also, $\tan(\text{A}+\text{B})=\sqrt{3}$
$\Rightarrow\tan(\text{A}+\text{B})=\tan60^\circ$ $\Big[\because\ \tan60^\circ=\sqrt{3}\Big]$
$\Rightarrow\text{A}+\text{B}=60^\circ\dots(\text{ii})$
Solving (i) and (ii), we get:
A = 45° and B = 15°.

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