Substitute the given values into the formula you found in the previous part.
E
e
Try to express h in terms of another angle measure using trigonometry.
A
a
See solution.
B
b
h=ABsin B
C
c
A=1/2(BC)ABsin B
D
d
A≈57.2 units^2
E
e
Example Solution: A=1/2(BC)ACsin C
Practice makes perfect
a
We are asked to draw an acute, scalene triangle ABC including an altitude of length h originating at vertex A. Let's do this. Remember that an altitude in a triangle is perpendicular to the base.
b
In this part we are asked to use trigonometry to express h in terms of m∠B. To do this, let's recall that in a right triangle the sine of an angle is the ratio of a side length opposite this angle to the length of the hypotenuse. Having this definition in mind, let's look at our triangle.
Since an altitude divided â–³ ABC into two right triangles, we can write an equation for sin B. The leg opposite to this angle has a length of h and the hypotenuse is AB.
sin B=h/AB
Next, we will isolate h in the above equation.
Now we will write an equation to find the area of a triangle ABC. Let's recall the formula for the area of a triangle.
A=1/2bh
In this formula, b is the length of the base and h is the length of an altitude of a triangle. In â–³ ABC, the altitude has a length of h, and the length of the base is BC.
Therefore, we can write an equation for the area of â–³ ABC.
A=1/2(BC) h
Since we are asked to use trigonometry in our equation, we will use the fact that we found that h=ABsin B.
In this part we are given that m∠B is 47^(∘), AB= 11.1, BC= 14.1, and CA=10.4 and asked to evaluate the area of △ ABC. To do this, we can use the formula we found in Part C.
To write an equation for the area of â–³ ABC using trigonometry in terms of a different angle measure, let's look at the picture for the last time.
First, we can express h in terms of m∠C. Again, we will use the definition of sine.
sin C=h/AC ⇓
h=ACsin C
Next, we will substitute the value of h into the area formula.
