Question

In triangle ABC, right angle at B, if AB = 12 cm and BC = 5 cm, find
1) Sin A and Tan A, 2) Sin C and Cot C.

Answer 1

This is your answer I hope this help you

Answer 2

The value of $\bold{\sin A=\frac{5}{13}, \ {Tan} A=\frac{5}{12}, \sin C=\frac{12}{13}, \quad \cot C=\frac{5}{13}}$.

Solution:

The triangle ABC is drawn below which is  

In the triangle ABC angle B is 90 degree. The “length of the side AB” = 12cm and the “length of the side BC” = 5cm.

Now to find the “length of side AC” we use Pythagoras theorem we get $A B^{2}+B C^{2}=A C^{2}$

$\sqrt{A B^{2}+B C^{2}}=A C$

$=\sqrt{12^{2}+5^{2}}$

$=\sqrt{144+25}=13$

Now the question says to find 1) Sin A and Tan A, 2) Sin C and Cot C.

So using the formula, we get

$\sin A=\frac{\text {height}}{\text {hypotenuse}}=\frac{B C}{A C}=\frac{5}{13}$

$\tan A=\frac{\text {height}}{\text {base}}=\frac{B C}{A B}=\frac{5}{12}$

$\sin C=\frac{\text {height}}{\text {hypotenuse}}=\frac{A B}{A C}=\frac{12}{13}$

$\cot C=\frac{\text {base}}{\text {height}}=\frac{B C}{A B}=\frac{5}{13}$

Therefore, the value of $\sin A=\frac{5}{13}, \ {Tan} A=\frac{5}{12}, \sin C=\frac{12}{13}, \quad \cot C=\frac{5}{13}$