Two sides of a triangle and their included angles are 4cm, 5cm and 30d

HI IanStewart,
What am I missing ? If the height is 2 and the base is 5 and its a right Triangle, then Hypotenuse should be
\(\sqrt{(2^2 +5^2)}\) =\(\sqrt{29}\) but here we have 4. How ?

Two sides of a triangle and their included angles are 4cm, 5cm and 30 degree resp. What is the are
What do you want to ask?

Solution:

The area of a triangle with sides a and b and the included angle C is Area = ½ ab sin C, where sin C means “the sine of angle C.” Therefore, the area of the triangle in this problem is:

Area = ½ * 4 * 5 * sin 30 = ½ * 20 * ½ = 5

Of course, using this formula requires us to have a basic understanding of trigonometry (for example, sin 30 = ½). However, this problem can be solved without any knowledge of trigonometry. This is how:

We can let 5 be the base of the triangle and 4 be the slant side that meets the base at a 30-degree angle. Now, if we construct the height of the triangle from the endpoint of the slant side (that doesn’t meet the base) to the base, we would have a right triangle with the slant side as the hypotenuse. Furthermore, this right triangle is a 30-60-90 right triangle with the height opposite the 30-degree angle, which makes the height ½ the length of the hypotenuse (i.e., the slant side). So the height is 2 for the right triangle. However, this is also the height of the overall triangle, i.e., the triangle with sides 5 and 4 and the included angle of 30 degrees. Since the overall triangle has a base of 5 and a height of 2, then its area is ½ x 5 x 2 = 5.

Answer: C

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