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In triangle ABC, a=6, b=7, c=10?

In triangle ABC, with side lengths a = 6, b = 7, and c = 10, what is the measure of angle A to the nearest tenth of a degree? Additionally, what is the area of triangle ABC, also rounded to the nearest tenth?

5 Answers

A
Anonymous

Dec 13, 2024

a)

With the Law of Cosines, we have:

a^2 = b^2 + c^2 – 2bc*cos(A).

With a = 6, b = 7, and c = 10, we have:

6^2 = 7^2 + 10^2 – 2(7)(10)*cos(A)

==> 36 = 49 + 100 – 140cos(A)

==> -140cos(A) = -113

==> cos(A) = 113/140

==> A ≈ 36°.

b)

From here, we can find that the area is:

Area = (1/2)bc*cos(A) = (1/2)(7)(10)cos(36°) ≈ 28.3 square units.

I hope this helps!

A
Anonymous

Jan 15, 2025

because of the fact the shortest edge of ABC is 24, the courting between 24 and six A’ is a million/4A consequently, because of the fact the triangle is similiar, A’B’C’ has all lengths a million/4 of the respective ABC lengths. consequently, longest part is going to be CA, at a million/4*40 = 10

A
Anonymous

Jan 29, 2025

Umm, your quesition is confusing. Are you giving me the angle measurements or the side lengths.

Since i dont know i’ll tell you the formulas you can use:

-to find area. 1/2 X Base of the triangle X the height of the triangle

to find degrees, you have to use SIN COS or TAN.

Sin is Opposite/hypotenuse side from the angle. Cos is Adjacent/ hypotenuse and Tan is Opposite/Adjacent

hope this helps a lil

A
Anonymous

Nov 25, 2024

use the cosine rule

Cos A = c^2+b^2-C^2 / 2ac

this is for the area

the areas half base times the height

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