Question

Class 10 Trigonometry pls fast i m in trouble......

Answer 1

Answer:-

45°

Explanation:-

Given

$tanA= \dfrac{1}{2}$

$tanB = \dfrac{1}{3}$

$tan(A+B) = \dfrac{tanA + tanB}{1-tanA \: tanB}$

To Find

A+B

Solution

we have

$tanA= \dfrac{1}{2}$

$tanB = \dfrac{1}{3}$

On putting the values

$tan(A+B) = \dfrac{\dfrac{1}{2} +\dfrac{1}{3} }{1-(\dfrac{1}{2})\:( \dfrac{1}{3})}$

$tan(A+B) = \dfrac{\dfrac{3+2}{6} }{1-\dfrac{1}{6}}$

$tan(A+B) = \dfrac{\dfrac{5}{6} }{\dfrac{6-1}{6}}$

$tan(A+B) = \dfrac{\dfrac{5}{6} }{\dfrac{5}{6}}$

$tan(A+B) = \dfrac{\cancel{\dfrac{5}{6} }}{\cancel{\dfrac{5}{6}}}$

$tan(A+B) = 1$

$\therefore A+B = 45 \degree$

Answer 2

$\huge\sf\underline{♡Answers♡}$

45°

Explanation—

Given that

$tanA= \dfrac{1}{2}$

$tanB = \dfrac{1}{3}$

$tan(A+B) = \dfrac{tanA + tanB}{1-tanA \: tanB}$

To Find—

A+B =?

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HERE UR EASY AND CATCHY SOLUTIONS—

we have

$tanA= \dfrac{1}{2}$

$tanB = \dfrac{1}{3}$

By substituting the values

$tan(A+B) = \dfrac{\dfrac{1}{2} +\dfrac{1}{3} }{1-(\dfrac{1}{2})\:( \dfrac{1}{3})}$

$tan(A+B) = \dfrac{\dfrac{3+2}{6} }{1-\dfrac{1}{6}}$

$tan(A+B) = \dfrac{\dfrac{5}{6} }{\dfrac{6-1}{6}}$

$tan(A+B) = \dfrac{\dfrac{5}{6}}{\dfrac{5}{6}}$

tan(A+B) = 1

Hence, A+B = 45°