Question

In the figure AC=CD,AD=BD and angle C=58° find ANGLE CAB

Answer 1

your ans is here
see the attachment!

Answer 2

Answer:

91.5° is the required value of ∠CAB

Step-by-step explanation:

Explanation:

Given , ABC is a triangle in which

AC = CD , AD = BD and  ∠C = 58°.

Therefore , AC = CD  and AD = BD

So , know that if two sides of a triangle are equal, then the angles opposite those sides are equal

⇒∠CAD = ∠CDA  and ∠ BAD = ∠DAB

Let ∠CAD = ∠CDA  be x and

Let ∠ BAD = ∠DAB be y

Step 1:

In ΔACD , ∠ACD = 58°   given

∠CAD + ∠CDA +∠ACD = 180          [sum of angle of triangle is 180  ]

⇒x + x + 58 = 180

⇒2x = 180- 58 = 122

⇒x = $\frac{122}{2}$ = 61

Therefore , ∠CAD = ∠CDA = 61 °

And ∠ ADC + ∠ADB = 180                    [Linear pair]

⇒ 61 + ∠ADB = 180                        [∠CDA =∠ADC = 61 ]

⇒∠ADB = 180  -61 = 119

Step 2:

In Δ ADB ,∠ADB = 119 .

∠ADB +∠ABD +∠DAB = 180     [Sum of angle of triangle is 180°]

⇒119 + y + y = 180

⇒119 + 2y = 180

⇒y = $\frac{180 - 119}{2} = \frac{61}{2}$= 30.5°

Therefore , ∠ BAD = ∠DAB = 30.5°

Step 3:

From step 1 and step 2 we get ,

∠ CAD = 61 °and ∠BAD = 30.5°

Therefore, ∠CAB = ∠ CAD + ∠BAD

⇒ ∠CAB = 61 + 30.5 = 91.5°

Final answer:

Hence ,  the angle of ∠CAB  is  91.5°.

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