Question

solve the question seen in the attachment

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Answer 1

Solution :-

SSS congruence rule = side - side - side

• if all three sides of two triangles are equal to each other, then the traingles are congruent by SSS congruency rule

CPCT : Corresponding parts of congruent triangles

• two or more triangles are congruent, then all of their corresponding angles and sides are congruent as well

__________________________

(i) prove that ∆ABM ≅ ∆ACM

in ∆ABM and ∆ACM

• AB = AC (given)
• BM = CM (given)
• AM = AM (common)

so, ∆ABM ≅ ∆ACM by SSS congruency rule

_______________________

(ii) find ∠CAM

we have, ∆ABM ≅ ∆ACM

➪ ∠CAM =∠BAM (by CPCT)

$\bold{: \implies \boxed{ \orange{\bold{∠CAM = 25 \degree}}}}$

_______________________

(iii) find ∠ABM

we have, ∆ABM ≅ ∆ACM

➪ ∠ABM =∠ACM (by CPCT)

$: \implies \boxed{ \orange{\bold{∠ABM = 65 \degree }}}$

________________________

(iv) Find ∠AMC

now, in ∆AMC

★ by using angles sum property of a triangle

• sum of all interior angles of a triangle is equal to 180°

$\bold{ : \implies ∠CAM + ∠ACM + ∠AMC = 180 \degree } \\ \\ \bold{: \implies25 \degree + 65 \degree +∠AMC = 180 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies 90 \degree +∠AMC = 180 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies ∠AMC = 180 \degree - 90 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \boxed{ \orange{ \bold{∠AMC = 90 \degree}}} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 2

Answer:

SSS congruence rule = side - side - side

if all three sides of two triangles are equal to each other, then the traingles are congruent by SSS congruency rule

CPCT : Corresponding parts of congruent triangles

two or more triangles are congruent, then all of their corresponding angles and sides are congruent as well

__________________________

(i) prove that ∆ABM ≅ ∆ACM

in ∆ABM and ∆ACM

AB = AC (given)

BM = CM (given)

AM = AM (common)

so, ∆ABM ≅ ∆ACM by SSS congruency rule

_______________________

(ii) find ∠CAM

we have, ∆ABM ≅ ∆ACM

➪ ∠CAM =∠BAM (by CPCT)

\bold{: \implies \boxed{ \orange{\bold{∠CAM = 25 \degree}}}}:⟹

∠CAM=25°

_______________________

(iii) find ∠ABM

we have, ∆ABM ≅ ∆ACM

➪ ∠ABM =∠ACM (by CPCT)

: \implies \boxed{ \orange{\bold{∠ABM = 65 \degree }}}:⟹

∠ABM=65°

________________________

(iv) Find ∠AMC

now, in ∆AMC

★ by using angles sum property of a triangle

sum of all interior angles of a triangle is equal to 180°

\begin{gathered} \bold{ : \implies ∠CAM + ∠ACM + ∠AMC = 180 \degree } \\ \\ \bold{: \implies25 \degree + 65 \degree +∠AMC = 180 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies 90 \degree +∠AMC = 180 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies ∠AMC = 180 \degree - 90 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \boxed{ \orange{ \bold{∠AMC = 90 \degree}}} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}

:⟹∠CAM+∠ACM+∠AMC=180°

:⟹25°+65°+∠AMC=180°

:⟹90°+∠AMC=180°

:⟹∠AMC=180°−90°

:⟹

∠AMC=90°