Question

Nᴇᴇᴅ Hᴇʟᴘ.ღ

$\huge{\mathfrak{\color{pink}{Qᴜᴇsᴛɪᴏɴ}}}$

Fɪɴᴅ Tʜᴇ Vᴀʟᴜᴇ Oғ "x" Fʀᴏᴍ Gɪᴠᴇɴ Fɪɢᴜʀᴇ...

NOTE:-
⇛nσ spαm
⇛nσ írrєlєvєnt αnswєr
⇛gívє prσpєr єхplαnαtíσn..
⇛αnswєr σf thє quєstíσn ís 90°
​

Answer 1

★ Concept :-

Here the concept of properties of Triangle has been used. We see that we are given a figure where three sides are equal. We need to find an angle. Firstly we will find the relation between all the angles of the triangle by using properties of Triangle. And then equate all the angles by using Angle Sum Property of Triangle to find the triangle.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{Sum\;of\;Angles\;of\;\Delta\;=\;180^{\circ}}}}$

______________________________________

★ Solution :-

Given,

» Triangle = ∆BRM

» Sub triangles = ∆BHM and ∆RHM

» BH = HM = RH

» ∠ M = x

» ∠x = ∠HMB + ∠HMR

» HM ⊥ BR

» ∠BHM = ∠RHM = 90°

• Theorem I :: If the opposite sides of triangle are equal, then the angle subtended by the equal sides are equal. This is the Equal and Opposite side property.

• Theorem II :: This states that the sum of all angles of triangle is equal to 180°. This is also called as the Angle Sum Property of Triangle.

-----------------------------------------------------------

~ For relationship between the different angles of ∆ ::

• Let's firstly consider ∆BHM ::

Here we see that BH = HM. So according to the Theorem I,

→ ∠B = ∠HMB ...(i)

• Let's now consider ∆RHM ::

Here we see that HM = HR. So according to the Theorem I,

→ ∠R = ∠HMR ...(ii)

• Let's now consider ∆BHM and ∆RHM ::

Here,

→ BH = RH

→ HM = HM (common sides)

→ ∠BHM = ∠RHM = 90° (since MH ⊥ BR)

So by SAS (side - angle - side) Conguruency,

→ ∆BHM ≈ ∆RHM

Thus, ∠HBM = ∠HRM by CPCT (Common Parts of Congruent Triangle)

So,

→ ∠B = ∠R ...(iii)

-----------------------------------------------------------

~ For the value of x ::

By Angle Sum Property of ∆ we know that,

$\;\sf{\rightarrow\;\;Sum\;of\;Angles\;of\;\Delta\;=\;180^{\circ}}$

By applying this in ∆BRM we get,

$\;\bf{\rightarrow\;\;\angle B\;+\;\angle M\;+\;\angle R\;=\;180^{\circ}}$

By applying values, we get

$\;\bf{\rightarrow\;\;\angle B\;+\;\angle x\;+\;\angle R\;=\;180^{\circ}}$

By applying equation (iii) we get,

$\;\bf{\rightarrow\;\;\angle B\;+\;\angle x\;+\;\angle B\;=\;180^{\circ}}$

(since ∠B = ∠R)

By applying the value of x here, we get

$\;\bf{\rightarrow\;\;\angle B\;+\;\angle HMB\;+\;\angle HMR \;+\;\angle B\;=\;180^{\circ}}$

From equation (i), equation (ii) and equation (iii) we get,

• ∠B = ∠HMB

• ∠B = ∠R

• ∠R = ∠HMR

From this we get,

• ∠B = ∠R = ∠HMR = ∠HMB

Now all these angles are equal here. So,

$\;\bf{\rightarrow\;\;4\angle HMB\;=\;180^{\circ}}$

$\;\bf{\rightarrow\;\;\angle HMB\;=\;\dfrac{180^{\circ}}{4}}$

$\;\bf{\rightarrow\;\;\angle HMB\;=\;\red{45^{\circ}}}$

From above relation we get,

→ ∠HMR = ∠HMB = 45°

And we also know that,

→ ∠x = ∠HMR + ∠HMB

→ ∠x = 45° + 45°

→ ∠x = 90°

This is the required answer.

$\;\underline{\boxed{\tt{Value\;\:of\;\:\angle x\;=\;\bf{\purple{90^{\circ}}}}}}$

______________________________________

★ More to know :-

• Equilateral Triangle : Triangle whose all the sides are equal and each angle is equal to 60°.

• Isoceles Triangle : Triangle whose two sides are equal.

• Scalene Triangle : Triangle whose all the sides are of different lengths.

• Acute - angled Triangle : Triangle whose all the angles measures to 90°

• Right - angled Triangle : Triangle whose anyone angle is equal to 90°.

• Obtuse - angled Triangle : Triangle whose one angle is more than 90°.

Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Concept}}}}$

Here in this question, concept of angle sum property as well as isosceles triangle is used. We see that in the given figure 3 sides are equal which will further prove both triangles (∆BMH & ∆MHR) as isosceles triangle. After providing both triangles as isosceles, we can easily find value of x. We have many methods to solve a particular question. Let's solve it by easiest method!!

Here we will use a theorem, which states:

If two sides of a triangle are equal, then angles opposite to sides will also be equal.

From this theorem:

• In ∆ BMH, BM is given to be equal to BH, so ∠1 will also be equal to ∠HMB...(1)

• In ∆ HRM, HR is given to be equal to HM, so ∠2 will also be equal to ∠RMH....(2)

In ∆RMB, by applying angle sum property:

⇒   ∠B+∠M+∠R=180°

We can notice from figure that ∠M comprises ∠HMB & ∠RMH.

⇒   ∠1+∠HMB+∠RMH+∠2=180°.

Now substituting values from (1) and (2).

⇒   ∠HMB+∠HMB+∠RMH+∠RMH=180°

⇒   2∠HMB+2∠RMH=180°

Now take 2 as common

⇒   2(∠HMB+∠RMH)=180°

Now transport 2 from LHS to RHS.

⇒   ∠HMB+∠RMH=180°/2

⇒   ∠HMB+∠RMH=90°

Now put ∠HMB+∠RMH=x

⇒    x=90°

So the required value of x is 90°.