Question

In the isosceles triangle ABC, angle A and Angle B are equal. angle ACB and angleACD are (3x + 17) and (8x + 10) degree respectively. Also find the measures of angle A and angle B.

Answer 1

★ Question:

In the isosceles ΔABC, ∠A, and ∠B are equal. ∠ACB and ∠ACD are (3x + 17)° and (8x + 10)° respectively. Also, find the measures of ∠A and ∠B.

★ To Find:

$\implies$  (3x + 17)° & (8x + 10)°

$\implies$ Measures of = ∠A and ∠B

★ Solution:

$\implies \sf{In \; the \; isosceles \; \triangle ABC, \; m\angle A = m\angle B } \dots \Big \langle \frak{\blue{Given}} \Big \rangle$

$\implies \sf{ \angle ACB \; and \; \angle ACD \; are \; (3x + 17)^\circ \; and \; (8x + 10)^\circ } \dots \Big \langle \frak{\blue{Given}} \Big \rangle$

$\implies \sf{m\angle ACD + m\angle ACB = 180^\circ} \dots \Big \langle \frak{\blue{Angles \; in \; a \; linear \; pair}} \Big \rangle$

$\implies \sf{(8x + 10)^\circ + (3x - 17)^\circ = 180^\circ}$

$\implies \sf{8x + 10 + 3x - 17 = 180^\circ}$

$\implies \sf{11x - 7 = 180^\circ}$

$\implies \sf{11x = 180 + 7}$

$\implies \sf{11x = 187}$

$\implies \sf{x = \dfrac{\cancel{187}}{\cancel{17}}}$

$\implies \underline{\boxed{\sf{\orange{x = 17}}}}$

   

m∠ACB,

$\implies \sf{3x - 17}$

$\implies \sf{3 \times 17 - 17}$

$\implies \sf{51 - 17}$

$\implies \underline{\boxed{\sf{\orange{m \angle ACB = 34^\circ}}}}$

   

m∠ACD,

$\implies \sf{8x + 10}$

$\implies \sf{8 \times 17 + 10}$

$\implies \sf{136 + 10}$

$\implies \underline{\boxed{\sf{\orange{m \angle ACD = 146^\circ}}}}$

   

By property,

The measures of an exterior angle of a triangle is equal to the sum of the measure of its remote interior angles.

$\implies \sf{m \angle ACD = m \angle A + m \angle B}$

$\implies \sf{146 = m \angle A + m \angle A \dots \Big \langle \frak{\blue{It \; is \; given \; that \; m \angle A = m\angle B}} \Big \rangle}$

$\implies \sf{146 = 2\; m \angle A}$

$\implies \sf{\dfrac{\cancel{146}}{\cancel{2}} = 73^\circ}$

$\implies \underline{\boxed{\sf{\orange{m \angle A = 73^\circ}}}}$

$\implies \underline{\boxed{\sf{\orange{m \angle B = 73^\circ}}}}$

   

★ Final Answer:

$\sf{\therefore \; the \; measure \; of \; the \; angles \; \angle ACB, \angle ACD, \angle A \; and \; \angle B \; are: } \\\underline{\sf{\; 34^\circ, \; 146^\circ, \; 73^\circ \; and \; 73^\circ \; respectively.}}$

Answer 2

Answer:

m<ACB = 34°

m<ACD = 146°

m<A = 73°

m<B =73°

Step-by-step explanation:

m<ACD is an exterior angle of ΔABC

m<ACD + m<ACB - Angles in a linear pair

$(8x + 10) + (3x - 17) = 180 ^\circ \\ 8x + 10 + 3x - 17 = 180 ^\circ \\ 11x = 180 + 7 \\ 11x = \frac{187}{11} \\ 11x = 17 \\ \boxed {x= 17}$

m<ACB ,

$\rightarrow \: 3x - 17 \\ </p><p>\rightarrow \: 3 x 17 - 7 \\ </p><p>\rightarrow \: 51 - 17$

: . m<ACB = 34°

m<ACD ,

$\rightarrow \: 8x + 10 \\ \rightarrow \: 8 \times 17 \\ \rightarrow \: 8 \times 17 + 10 \\ \rightarrow \: 136 + 10$

: . m<ACD = 146°