Question

In this triangle, the product of sin B and tan C is and the product of sin Cand tan B is

Answer 1

Answer:

$\frac{c}{a}$ and $\frac{b}{a}$

Step-by-step explanation:

sinB = $\frac{opposite}{hypotenuse}$ = $\frac{AC}{BC}$ = $\frac{b}{a}$

tanC = $\frac{opposite}{adjacent}$ = $\frac{AB}{AC}$ = $\frac{c}{b}$

Thus

sinB tanC = $\frac{b}{a}$ × $\frac{c}{b}$ ( cancel b on numerator/ denominator )

                 = $\frac{c}{a}$

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sinC = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{BC}$ = $\frac{c}{a}$

tanB = $\frac{opposite}{adjacent}$ = $\frac{AC}{AB}$ = $\frac{b}{c}$

Thus

sinC tanB = $\frac{c}{a}$ × $\frac{b}{c}$ ( cancel c on numerator/ denominator )

                 = $\frac{b}{a}$