Question

In triangle ABC ,angke b =90° , BC=5cm, AC- AB=1cm . Evaluate 1+sin c/ cos c

Answer 1

Answer:

The value of $\frac{1+sinC}{cosC}$ is

5

Step-by-step explanation:

Given:

∠B = 90°

BC = 5 cm

AC - AB = 1 cm

Take AB = x

then AC = x+1

In right ΔABC,

$AC^2=AB^2+BC^2$

$(x+1)^2=x^2+5^2$

$x^2+1+2x=x^2+25$

$1+2x=25$

$2x=24$

$x=12\:cm$

AB=12 cm, AC=13 cm

$sin\:C=\frac{AB}{AC}$

$sin\:C=\frac{12}{13}$

$cos\:C=\frac{BC}{AC}$

$cos\:C=\frac{5}{13}$

Now,

$\frac{1+sinC}{cosC}$

$=\frac{1+\frac{12}{13}}{\frac{5}{13}}$

$=\frac{\frac{13+12}{13}}{\frac{5}{13}}$

$=\frac{\frac{25}{13}}{\frac{5}{13}}$

$=\frac{25}{5}$

$=5$

Therefore, the value of $\frac{1+sinC}{cosC}$ is 5

Answer 2

HOPE IT HELPS YOU....

@RUKSHI...