Question

ΔABC is a right triangle in which ∠B is a right angle, AB = 1, AC = 2, and BC = sqrt(3).

cos C × sin A =

I think the answer is 3/4, because cos(c) = adj / hypot and sin(a) = opposite / hypot

cos(c) = sqrt(3) / 2
sin(a) = sqrt(3) / 2
which is 3/4 when multiplied.

Answer 1

Answer:

$\frac{3}{4}$

Step-by-step explanation:

cosC = cos30° = $\frac{adjacent}{hypotenuse}$ = $\frac{BC}{AC}$ = $\frac{\sqrt{3} }{2}$

sinA = sin60° = cos30° = $\frac{\sqrt{3} }{2}$

Hence

cosC × sinA = $\frac{\sqrt{3} }{2}$  × $\frac{\sqrt{3} }{2}$ = $\frac{3}{4}$

Answer 2

Answer:

$\frac{3}{4}$

Step-by-step explanation:

Since,

$\sin \theta=\frac{Opposite leg of }\theta}{\text{Hypotenuse}}$

$\cos \theta=\frac{Adjacent leg of }\theta}{\text{Hypotenuse}}$

Given,

In triangle ABC,

AB = 1 unit, AC = 2 unit, and BC = √3 unit

Thus, by the above formule,

$\cos C = \frac{\sqrt{3}}{2}$

$\sin A=\frac{\sqrt{3}}{2}$

$\implies \cos C\times \sin A =  \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}= \frac{3}{4}$