Michael Friesen
Dec 27, 2024
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)
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!
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
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
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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