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9.Straight Line
normal
રેખાઓ $3x + y + 4 = 0$ , $3x + 4y -15 = 0$ અને $24x -7y = 3$ થી ..............ત્રિકોણ બને
A
સમબાજુ
B
સમદ્રીબાજુ
C
લઘુકોણ
D
વિષમકોણ
Solution
$\frac{24}{7}>-\frac{3}{4}>-3$
$\tan A=\frac{\frac{24}{7}+\frac{3}{4}}{1-\frac{24}{7} \cdot \frac{3}{4}}=-\frac{117}{51},$
$\tan B=\frac{-\frac{3}{4}+3}{1+\frac{9}{4}}=\frac{9}{13}$
$\tan C=\frac{-3-\frac{24}{7}}{1-3 \cdot \frac{24}{7}}=\frac{45}{65}=\frac{9}{13}$
Standard 11
Mathematics
