SOLUTION: In triangle ABC, (bsin C) (bcos C + ccos B) =42 Compute the area of the triangle.

Question 1192620: In triangle ABC, (bsin C) (bcos C + ccos B) =42
Compute the area of the triangle.
Found 3 solutions by ikleyn, MathLover1, math_tutor2020:
Answer by ikleyn(52905)   (Show Source): You can put this solution on YOUR website!
.

Make a sketch. b*cos(C) + c*cos(B) is the length of the side "a" of the triangle ABC. b*sin(C) is the length of the altitude drawn from vertex A to side "a". Therefore, the area of the triangle ABC is half of the product of the two expressions, i.e. 42/2 = 21 square units. ANSWER

Solved.

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Regarding the post by @MathLover1, this lady makes a lot (tons) of UNNECESSARY calculations,

while the problem's solution is MUCH EASIER   and   MUCH SIMPLER,   as shown in my post.

The meaning of this assignment is not manipulating formulas.

Its meaning is to discern geometric meaning of each term of the formula directly from the formula.

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

given:

is the length of the side of the triangle ABC

use the law of cosine to prove it:

then we have

=
=
=
=
=

then we have
is the length of the altitude drawn from vertex A to the side , and the area of triangle is:

sq. units

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Here's one way to draw out what the tutor @ikleyn is referring to

Useful things to memorize:
sin(angle) = opposite/hypotenuse
cos(angle) = adjacent/hypotenuse
These two formulas apply to right triangles only.

Further note:
b*sin(C) = c*sin(B) as they both represent the purple height segment.

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