The roof on a house has one side that is in the shape of an isosceles triangle. If the sides of this part are 18 feet long and the angle at the peak is 50°, what is the area of this part of the roof to the nearest tenth of a foot?

124.1 feet

Step-by-step explanation:

An isosceles triangle is one which has two equal sides and two equal angles.

Area of triangle = ab SinC

where a and b are the sides, and C is the given included angle.

From the given question, a = b = 18 feet and C = 50°.

Then,

Area of this part of the roof = x 18 x 18 x Sin 50°

                                                 = x 18 x 18 x 0.7660

                                                = 162 x 0.7660

                                                 = 124.092

The area of this part of the roof is 124.1 feet.

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb