The sides a, b, c of a triangle satisfy the r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given relations for the sides of the triangle are:$c^2 = 2ab$  $(i)$$a^2 + c^2 = 3b^2$  $(ii)$Substituting $(i)$ into $(ii)$:$a^2 + 2ab = 3b^2$$a^

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given relations for the sides of the triangle are:$c^2 = 2ab$  $(i)$$a^2 + c^2 = 3b^2$  $(ii)$Substituting $(i)$ into $(ii)$:$a^2 + 2ab = 3b^2$$a^2 + 2ab - 3b^2 = 0$$(a + 3b)(a - b) = 0$Since $a$ and $b$ are side lengths,$a, b > 0$,so $a = b$.Substituting $a = b$ into $(i)$:$c^2 = 2(b)(b) = 2b^2$$c = \sqrt{2}b$Now,we have sides $a = b$,$b = b$,and $c = \sqrt{2}b$.Since $a^2 + b^2 = b^2 + b^2 = 2b^2 = c^2$,the triangle is a right-angled triangle with the right angle at $C$ (i.e.,$\angle C = 90^{\circ}$).Since $a = b$,the triangle is an isosceles right-angled triangle.Therefore,$\angle A = \angle B = 45^{\circ}$.The measure of $\angle BAC$ is $45^{\circ}$.

The sides $a, b, c$ of a triangle satisfy the relations $c^2=2ab$ and $a^2+c^2=3b^2$. Then the measure of $\angle BAC$,in degrees,is

Similar Questions

In $\triangle ABC$,find the value of $a^3 \cos(B-C) + b^3 \cos(C-A) + c^3 \cos(A-B)$.

The equation $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$ has a solution if

In a right-angled triangle,if the difference between the two acute angles is $60^{\circ}$,then the ratio of the length of the hypotenuse to the length of the perpendicular drawn to the hypotenuse from its opposite vertex is: (in $: 1$)

Let $PQR$ be a triangle of area $\Delta$ with $a=2, b=\frac{7}{2}$ and $c=\frac{5}{2}$,where $a, b$ and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 \sin P-\sin 2P}{2 \sin P+\sin 2P}$ equals

In triangle $ABC$,$\frac{\tan A}{2} = \frac{\tan B}{3} = \frac{\tan C}{4}$,then the value of $\sec^2 A + \sec^2 B + \sec^2 C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module