Advertisements
Advertisements
Question
Derive Projection formula from Law of cosines
Advertisements
Solution
To prove
(a) a = b cos C + c cos B
(b) b = c cos A + a cos C
(c) c = a cos B + b cos A
Using Law of cosines.
We have cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`
cos B = `("c"^2 + "a"^2 - "b"^2)/(2"ca")`
cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
(a) b cos C + cs B
= `"b"(("a"^2 + "b"^2 - "c"^2)/(2"ab")) + "c"(("c"^2 + "a"^2 - "b"^2)/(2"ca"))`
= `("a"^2 + "b"^2 - "c"^2)/(2"a") + ("c"^2 + "a"^2 - "b"^2)/(2"a")`
= `("a" + "b"^2 - "c"^2 + "c"^2 + "a"^2 - "b"^2)/(2"a")`
= `(2"a"^2)/(2"a")`
= a
∴ a = b cos C + c cos B
(b) c cos A + a cos C
= `"c"(("b"^2 + "c"^2 - "a"^2)/(2"bc")) + "a"(("a"^2 + "b"^2 - "c"^2)/(2"ab"))`
= `("b"^2 + "c"^2 - "a"^2)/(2"bc") + ("a"^2 + "b"^2 - "c"^2)/(2"ab")`
= `("b"^2 + "c"^2 - "a"^2 + "a"^2 + "b"^2 - "c"^2)/(2"b")`
= `(2"b"^2)/(2"b")`
= b
∴ b = c cos A + a cos C
(c) c = a cos B + b cos A
= `"a"(("c"^2 + "a"^2 - "b"^2)/(2"ca")) + "b"(("b"^2 + "c"^2 - "a"^2)/(2"bc"))`
= `("c"^2 + "a"^2 - "b"^2)/(2"c") + ("b"^2 + "c"^2 - "a"^2)/(2"c")`
= `("c"^2 + "a"^2 - "b"^2 + "b"^2 + "c"^2 - "a"^2)/(2"c")`
= `(2"c"^2)/(2"c")`
= c
∴ c = a cos B + b cos A
APPEARS IN
RELATED QUESTIONS
In a ∆ABC, if `sin"A"/sin"C" = (sin("A" - "B"))/(sin("B" - "C"))` prove that a2, b2, C2 are in Arithmetic Progression
The angles of a triangle ABC, are in Arithmetic Progression and if b : c = `sqrt(3) : sqrt(2)`, find ∠A
In a ∆ABC, prove that `sin "B"/sin "C" = ("c" - "a"cos "B")/("b" - "a" cos"C")`
In an ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C
In a ∆ABC, ∠A = 60°. Prove that b + c = `2"a" cos (("B" - "C")/2)`
In an ∆ABC, prove the following, `"a"sin ("A"/2 + "B") = ("b" + "c") sin "A"/2`
In a ∆ABC, prove the following, a(cos B + cos C) = `2("b" + "c") sin^2 "A"/2`
In a ∆ABC, prove the following, `("a"^2 - "c"^2)/"b"^2 = (sin ("A" - "C"))/(sin("A" + "C"))`
In a ∆ABC, prove the following, `("a"sin("B" - "C"))/("b"^2 - "c"^2) = ("b"sin("C" - "A"))/("c"^2 - "a"^2) = ("c"sin("A" - "B"))/("a"^2 - "b"^2)`
An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park
A rope of length 42 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed
Derive Projection formula from Law of sines
A circle touches two of the smaller sides of a ΔABC (a < b < c) and has its centre on the greatest side. Then the radius of the circle is ______.
In a ΔABC, let BC = 3. D is a point on BC such that BD = 2, Then the value of AB2 + 2AC2 – 3AD2 is ______.
In an equilateral triangle of side `2sqrt(3)` cm, the circum radius is ______.
If in a ΔABC, the altitudes from the vertices A, B, C on opposite sides are in H.P, then sin A, sin B, sin C are in ______
