Question 62

Ina $$\triangle$$ABC,right angled at B, AB = 7 cm and (AC - BC) = 1 cm. The value of $$(\sec C + \cot A)$$ is :

Solution

We have AC-BC =1
Now AC^2-BC^2=49
So (AC-BC((AC+BC) =49
So AC+BC =49
Solving we get AC=25 ;BC =24
secC +cotA = (25+7)/24 = 32/24 =4/3

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