If and ; ; , find and .

Understand the Question Given trigonometric values for compound angles: and . Using standard trigonometric values for acute angles (): Solving this system of two linear equations simultaneously gives the individual values of and . Step 1 · Form Linear Equations from Given Values Given Since Also given Since Step 2 · Solve for and Adding equations and Substitute into equation Answer > and Common Mistakes Distributive Error: Assuming . Trigonometric ratios cannot be distributed over angle sums or differences. Angle Value Swap: Mixing up standard values, such as writing and . Condition Check: Always verify the constraints given: satisfies , and satisfies .

Introduction to Trigonometry | Exercise 8.2

Question 3

If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \dfrac{1}{\sqrt{3}}$ ; $0^\circ < A + B \le 90^\circ$ ; $A > B$ , find $A$ and $B$ .

Solution

Understand the Question

Given trigonometric values for compound angles: $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \dfrac{1}{\sqrt{3}}$ .
Using standard trigonometric values for acute angles ( $\,0^\circ < A+B \le 90^\circ$ $0^{\circ} < A + B \leq 9 0^{\circ}$ ):
- $\tan 60^\circ = \sqrt{3} \implies A + B = 60^\circ$
- $\tan 30^\circ = \dfrac{1}{\sqrt{3}} \implies A - B = 30^\circ$
Solving this system of two linear equations simultaneously gives the individual values of $A$ and $B$ .

Step 1 · Form Linear Equations from Given Values

Given $\tan (A + B) = \sqrt{3}$

Since $\tan 60^\circ = \sqrt{3}$ $\tan (A + B) = \tan 60^\circ \implies A + B = 60^\circ \quad \dots (1)$

Also given $\tan (A - B) = \dfrac{1}{\sqrt{3}}$

Since $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ $\tan (A - B) = \tan 30^\circ \implies A - B = 30^\circ \quad \dots (2)$ Diagram 1

Step 2 · Solve for $A$ and $B$

Adding equations $(1)$ and $(2)$

\begin{aligned} (A + B) + (A - B) &= 60^\circ + 30^\circ \\[0.6em] 2A &= 90^\circ \\[0.6em] A &= \dfrac{90^\circ}{2} = 45^\circ \end{aligned}

Substitute $A = 45^\circ$ into equation $(1)$

\begin{aligned} 45^\circ + B &= 60^\circ \\[0.6em] B &= 60^\circ - 45^\circ = 15^\circ \end{aligned}

Answer

$A = 45^\circ$ and $B = 15^\circ$

Common Mistakes

Distributive Error: Assuming $\tan(A + B) = \tan A + \tan B$ . Trigonometric ratios cannot be distributed over angle sums or differences.
Angle Value Swap: Mixing up standard values, such as writing $\tan 30^\circ = \sqrt{3}$ and $\tan 60^\circ = \dfrac{1}{\sqrt{3}}$ .
Condition Check: Always verify the constraints given: $A + B = 60^\circ$ satisfies $0^\circ < A + B \le 90^\circ$ , and $A = 45^\circ > B = 15^\circ$ satisfies $A > B$ .

Question 3

More questions in Exercise 8.2