Question

solve the question seen in the attachment ​

Answer 1

Answer:

i. we know that

AB=AC{given}

BM=CM{given}

AM is common

ii. we know that

angle CAM+ angle ACM + angle AMC = 180°

angle CAM + 65° + 90° = 180°

angle CAM = 180-150

angle CAM = 30°

iii. we know that

25 + 90 + ABM = 180°

ABM = 180-145

ABM = 25

Answer 2

Answer:

i. Prove that ∆ABM = ∆ACM

In ∆ABM = ∆ACM,

AB = AC

Angle AMB = Angle AMC [ 90° ]

BM = MC

Hence, ∆ABM = ∆ACM is congruent by RHS Criterian.

ii. Find Angle CAM

Since both the triangles are congreunt so,

Hence, Angle MAB = Angle CAM = 25°

iii. Find Angle ABM

Since both the triangles are congreunt so,

Hence, Angle ACM = Angle ABM = 65°

iv. Find Angle AMC

Angle Sum Property of a Triangle = 180°

Angle CAM + Angle ACM + Angle AMC = 180°

=> 25° + 65° + Angle AMC = 180°

=> 90° + Angle AMC = 180°

=> Angle AMC = 90°

Another reason could be that since in the Case 1 we identified that the triangles are congruent by RHS Criterian, so directly we could have said that Angle AMC = 90°.