Question

the altitude of the triangle abc, segment da, has been drawn. derive a formula for the area of triangle bca that depends on the measure of angle b along with side lengths a and c. area = 1/2(base)(height) following this setup area = 1/2ah sin(b) = , so h = , and area =

Explanation:

Step1: Define sine in right - triangle

In right - triangle BDA, $\sin(B)=\frac{h}{c}$, because sine of an angle in a right - triangle is the ratio of the opposite side to the hypotenuse.

Step2: Solve for h

From $\sin(B)=\frac{h}{c}$, we can solve for $h$ by multiplying both sides of the equation by $c$. So, $h = c\sin(B)$.

Step3: Substitute h into area formula

The area of triangle BCA is $A=\frac{1}{2}ah$. Substitute $h = c\sin(B)$ into the area formula, we get $A=\frac{1}{2}ac\sin(B)$.

Answer:

$\sin(B)=\frac{h}{c}$, $h = c\sin(B)$, Area $=\frac{1}{2}ac\sin(B)$