Question

In a triangle ABC, right angled at A, if AB = 5, AC = 12 and BC = 13, find sinB, cosC and tanB.​

Answer 1

$\mathtt{\bf{\large{\underline {\red{Given \: Question}}}}}$

In a triangle ABC, right angled at A, if AB = 5, AC = 12 and BC = 13, find sinB, cosC and tanB.

$\mathtt{\bf{\large{\underline {\red{Required\: Answer}}}}}$

$★ \: sinB = \frac{12}{13} \\ \\ ★ \: tanB = \frac{12}{5} \\ \\★ \: cosC = \frac{12}{13}$

$\mathtt{\bf{\large{\underline {\red{Explanation:-}}}}}$

$\mathtt{\bf{\large{\underline {\blue{Given:-}}}}}$

AB = 5

AC = 12

BC = 13

$\mathtt{\bf{\large{\underline {\blue{To \: find:-}}}}}$

$☞ \: sinB \\☞ \: tanB \\ ☞ \: cosC$

$\mathtt{\bf{\large{\underline {\blue{Solution:-}}}}}$

$\color{purple}{With \: \: reference \: \: to \: ∠B, }$

• Perpendicular ⇨ CA⇨12

• Base ⇨AB⇨5

• Hypotenuse ⇨BC⇨13

$\color{purple}{So,}$

$⟾ \: \: sinB = \frac{perpendicular}{hypotenuse} = \frac{CA}{BC} = \frac{12}{13} \\ \\⟾ \: \: tanB= \frac{perpendicular}{base} = \frac{CA}{AB} = \frac{12}{5}$

$\color{purple}{With \: \: reference \: \: to \: \: ∠C, }$

• Perpendicular ⇨ BA⇨5

• Base ⇨CA ⇨ 12

• Hypotenuse ⇨CB⇨13

$\color{purple}{So, }$

$⟾ \: \: cosC = \frac{base}{hypotenuse} = \frac{CA}{CB} = \frac{12}{13}$

$\fbox{Additional Information}$

$➻ \: sin θ \: = \frac{perpendicular}{hypotenuse} \\ \\➻ \: cos θ = \frac{base }{hypotenuse} \\ \\ ➻ \: tan θ \: = \frac{perpendicular}{base}$

Answer 2

sin= 5/12

cos = 13/5

tan = 5/13