Advertisement Remove all ads

# In ∆ABC, if 2cosAa+cosBb+2cosCc=abc+bca, then show that the triangle is a right angled - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

In ∆ABC, if (2cos "A")/"a" + (cos "B")/"b" + (2cos"C")/"c" = "a"/"bc" + "b"/"ca", then show that the triangle is a right angled

Advertisement Remove all ads

#### Solution

In ΔABC by cosine rule, we get

cos A = ("b"^2 + "c"^2 - "a"^2)/(2"bc"), cos B = ("a"^2 + "c"^2 - "b"^2)/(2"ac"), cos C = ("a"^2 + "b"^2 - "c"^2)/(2"ab")

(2cos"A")/"a" + (cos "B")/"b" + (2cos"C")/"c" = "a"/"bc" + "b"/"ca"  .......[Given]

∴ (2("b"^2 + "c"- "a"^2))/(2"abc") + ("a"^2 + "c"^2 - "b"^2)/(2"abc") + (2("a"^2 + "b"^2 - "c"^2))/(2"abc") = (2"a"^2 + 2"b"^2)/(2"abc")

∴ 2b2 + 2c2 – 2a2 + a2 + c2 – b2 + 2a2 + 2b2 – 2c2 = 2a2 + 2b2

∴ b2 – a2 + c2 = 0

∴ a2 = b2 + c2

Hence, ΔABC is a right angled triangle.

Concept: Solutions of Triangle
Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?