In Delta ABC,m angle A = m angle B + m angle | vedclass

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

Question

(C) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A = m \angle B + m \angle C$,substitutin

Vedclass · Accepted Answer

$2$ or $6$

Vedclass · Answer

(C) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A = m \angle B + m \angle C$,substituting this into the sum gives $m \angle A + m \angle A = 180^{\circ}$,so $2m \angle A = 180^{\circ}$,which means $m \angle A = 90^{\circ}$.Since $\overline{AM}$ is an altitude to $\overline{BC}$,$\Delta ABM$ and $\Delta ACM$ are right-angled triangles at $M$.Let $BM = x$. Then $MC = 8 - x$.In $\Delta ABM$,$AB^2 = AM^2 + BM^2 = 12 + x^2$.In $\Delta ACM$,$AC^2 = AM^2 + MC^2 = 12 + (8 - x)^2$.In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + AC^2 = BC^2$,so $(12 + x^2) + (12 + (8 - x)^2) = 8^2$.$24 + x^2 + 64 - 16x + x^2 = 64$.$2x^2 - 16x + 24 = 0$.$x^2 - 8x + 12 = 0$.$(x - 6)(x - 2) = 0$.Thus,$x = 6$ or $x = 2$.

In $\Delta ABC$,$m \angle A = m \angle B + m \angle C$ and $\overline{AM}$ is an altitude. If $AM = \sqrt{12}$ and $BC = 8$,find $BM$.

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. If $BD = 2 \sqrt{30}$ and $CD = 6$,then $AC = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$,and $AM > CM$. If $BM = 12$ and $AC = 25$,find $AB$ and $BC$.

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and $BC = 18$,then the area of $\Delta ABC$ is.............

In $\Delta ABC$,if $\ldots \ldots \ldots \ldots$,then by Apollonius' theorem,$AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})$ holds good.

Vedclass Test Series

Exam Paper Generator

Online Exam Module