Question

In triangle ABC, if angle B = 90°, BC = 5 cm and AC - AB = 1, evaluate the value of $\frac{1 + \sin(c) }{1 + \cos(c) }$
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Answer 1

Answer:

in ∆ABC

BC = 5cm and AC - AB = 1 this implies AC = 1 + x

let AB = x

using Pythagoras theorem in ∆ABC

AB² + BC² = AC²

x² + 5² = (1+x)²

x² + 25 = 1 + x² + 2x

2x = 24

x = 12cm = AB

then AC = 12 + 1 = 13cm

(1+ sin C) / (1+cosC)

(1+ AB/AC)/(1+BC/AC)

(1 + 12/13) / (1+5/13)

(25/13)/(18/13)

25/18

hope this will help you

Answer 2

Answer:

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