TRIANGLE ANGLES TEST-1

TYT-AYT Geometry issues, angles in triangle test-1 and their solutions ..

Question 1-) If ABC is a triangle, m (DBA) = m (DCB), m (BDC) = 143 °; How many degrees is m (CBA) = x?

Question 2-) If ABC is a triangle, m (BAD) = m (DAC), m (CBA) – m (ACB) = 32 °; How many degrees is m (ADB)?

Question 3-) If ABC is a triangle, [AD] perpendicular [BD], m (DBA) = m (DAC), m (CBA) = 40 °; How many degrees is m (ACB) = x?

Question 4-) If ABC is a triangle, [BD] and [CD] bisector, m (BAC) = 2x, m (BDC) = 6x; How many degrees is x?

Question 5-) [BE parallel [CF, [AP] and [DP] bisector, if m (ABE) = 141 °, m (FCD) = 151 °; How many degrees is m (DPA) = x?

Question 6-) If ABC is a triangle, [AE] and [BD] bisector, m (BDC) = 115 °, m (AEB) = 76 °; How many degrees is m (ACB) = x?

Question 7-) If ABC is a triangle, [BD] and [CD] bisector, m (EAF) = 14x, m (CDB) = 11x; How many degrees is x?

Question 😎 ABC is a triangle, [BD] and [CD] bisector, if m (EBF) = 44 °; How many degrees is m (ADC) = x?

TEST-1 SOLUTIONS FOR TRIANGLE ANGLES

Solution 1: If the angle B is x, then the CBD angle is x-a. From the sum of the angles of the DBC triangle, x = 37 °

Solution 2: From the sum of the angles of the US triangle, x + n = 180 ° -m (ADB), in the ADC triangle from the outer angle y + n = m (ADB). Multiply the second expression by (-2) and add xy = 180 to the side by the first. ° -2m (ADB). If 32 ° is written instead of x, m (ADB) = 74 °.

Solution 3: The size of the angle formed when one side of the triangle is bent inward is x = a + b + c. The US concave quadrilateral contains x + a + 40 ° -a = 90 ° (shalwar, rocket, missile rule), x = 50 °. See geometry proofs section for proof of rule-Rocket rule-Missile rule and other proofs.

Solution 4: The size of the angle formed by the two inner bisectors in a triangle is 90 degrees more than half the size of the inside angle in the third corner, so 6x = 90 ° + x, x = 18 °.

Solution 5: Let the extension [AB] and [DC] intersect at point K. With the rule of M, m (DKA) = 68 °. The dimension of the angle formed by the intersection of two bisector in the triangle AKD is x = 90 ° + (68 ° / 2) = 124 ° .

Solution 6: Since [AE] and [BD] are bisectors in triangle ABC, m (AFB) = 90 ° + (x / 2). If the angles of the quadrant equal 360 °, then x = 34 °.

Solution 7: The measure of the angle formed by the two outer bisectors in a triangle is the complement of half the measure of the angle in the third corner. 11x = 90 ° – (14x / 2), x = 5 °.

Solution 8: The angle between the two outer bisectors is x = 90 ° – (44 ° / 2) = 68 °.