Question

Calculate the magnitude of the angle marked as x in each of the following triangles.

(a) Ans — angle P = 50°
(b) Ans — angle A = 60°
(c) Ans — angle A = 75°
(d) Ans — angle R = 110° ​

Answer 1

Step-by-step explanation:

1) Given: Triangle (a)

To find:

$\angle \: P$

Solution:

$\angle \: SQP + \angle \:PQR = 180° \: \: \: \: (linear \: pair) \: \\ \\ 120° + \angle \: PQR = 180° \\ \\ \angle \: PQR = 180° - 120° \\ \\ \angle \: PQR = 60°$

By the same way in the other side

$\angle \: PRT + \angle \: PRQ = 180° \: \: \: \: (linear \: pair) \: \\ \\ 110° + \angle \: PRQ = 180° \\ \\ \angle \: PRQ = 180° - 110°\\ \\ \angle \: PRQ = 70°$

In ∆PQR

Sum of all three angles is 180°

$\angle \: PRQ + \angle \: RPQ + \angle \: QPR = 180° \\ \\ 60° + 70° +\angle \: QPR = 180° \\ \\ \angle \: QPR = 180° - 130° \\ \\ \angle \: QPR = 50°$

Thus,

$\bold{\angle \: P= 50°}$

b) Given: Triangle (b)

To find:

$\angle \: A$

Solution:

$\angle \: DCA + \angle \:ACB = 180° \: \: \: \: (linear \: pair) \: \\ \\ 90° + \angle \: ACB = 180° \\ \\ \angle \: PQR = 180° - 90° \\ \\ \angle \: ACB = 90°$

In ∆ABC

Sum of all three angles is 180°

$\angle \: ABC + \angle \: ACB + \angle \: BAC = 180° \\ \\ 30° + 90° +\angle \: BAC = 180° \\ \\ \angle \: BAC = 180° - 120° \\ \\ \angle \: BAC = 60° \\ \\\bold{\angle\:A=60°}$

c)Given: Triangle (c)

To find:

$\angle \: A$

Solution:

$\angle \: ACD + \angle \:ACB = 180° \: \: \: \: (linear \: pair) \: \\ \\ 135° + \angle \: ACB = 180° \\ \\ \angle \: ACB = 180° - 135° \\ \\ \angle \: ACB = 45°$

In ∆ABC

Sum of all three angles is 180°

$\angle \: ABC + \angle \: ACB + \angle \: BAC = 180° \\ \\ 60° + 45° +\angle \: BAC = 180° \\ \\ \angle \: BAC = 180° - 105° \\ \\ \angle \: BAC = 75° \\ \\\bold{\angle\:A=75°}$

d) Given: Triangle (d)

To find:

$\angle \: R$

SPT is a straight line,sum of all angles is 180°

$\angle \: SPQ + \angle \: QPR + \angle \: TPR = 180° \\ \\ 60° + \angle \: QPR +70° = 180° \\ \\ \angle \: QPR = 180° - 130° \\ \\ \angle \: QPR = 50°$

EXTERIOR ANGLE OF TRIANGLE IS EQUAL TO SUM OF TWO OPPOSITE INTERIOR ANGLES OF TRIANGLE.

Thus,

$\angle \: PRU= \angle \: PQR + \angle \: QPT \\ \\ \angle \: PRU=60° + 50° \\ \\ \angle \: PRU = 110°\\ \\ \bold{\angle \: R = 110°}$

Hope it helps you.

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Answer 2

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