TRIANGLE ANGLES TEST-2

TYT-AYT Geometry issues, angles in triangle test-2 and their solutions …

Question 1-) ABC is a triangle, [BD] and [CD] bisector, if m (BDC) = 52 °; How many degrees is m (BAC) = x?

Question 2-) ABC is a triangle, [BD] and [CD] bisector, if m (BDC) = 50 °; How many degrees is m (CAD) = x?

Question 3-) In ABC triangle; D is the center of the outer tangent circle, if m (ADB) = x, m (BDC) = y, m (BAC) = 7x-5 °, m (ACB) = y-20 °; How many degrees is x?

Question 4-) If KN parallel [CP, m (ACB) = m (BCP), | AC | = | BC |, m (CAK) = 60 °; How many degrees is m (NAB) = x?

Question 5-) If [AD parallel NP, [AC] bisector, | AB | = | CD |, m (ABN) = 2m (PBD); How many degrees is m (DCA) = x?

Question 6-) ABC is a triangle, | AB | = | AE | = | BD |, if m (EAC) = 24 °, m (CBD) = 30 °; How many degrees is m (ACB) = x?

Question 7-) ABC triangle, if | AP | = | PC |, | FA | = | FB | = | FP |, if m (ACB) = 33 °; How many degrees is m (BAF) = x?

Question 😎 If | AB | = | BC | = | CD | = | DE |, if m (EDC) = m (CDB); How many degrees is m (DAE) = x?

TEST-2 SOLUTIONS FOR TRIANGLE ANGLES

Solution: m (BDC) = 52 °, formed by the intersection of triangle ABC with an inner bisector and an outer bisector, will be half the angle m (BAC), so m (BAC) = 2.52 ° = 104 °.

Solution: Since the triangle ABC intersects an inner bisector and an outer bisector, m (BAC) = 100 °. If there is any two of the two outer bisectors in the triangle, the third is the bisector. (Point D is the center of the outer tangent circle of triangle ABC.) In this case, m (CAD is CAD. ) = x = 40 °.

Solution: [AD], [BD], [CD] is bisector. ABC triangle; With the inner bisector of [BD], 2y = 7x-5 ° from the outer bisector of [CD], and 2x = y-20 ° from the outer bisector [AD] with the inner bisector of [BD].

Solution: Equilateral angles from inside angles are 30 degrees. In an isosceles triangle, m (BAC) = 75 °. From the right angle x = 45 °.

Solution: If we say m (BAC) = m (CAD) = a, then m (ABN) = 2a from inside angles. If m (ABN) = 2a then m (PBD) = a. M (ADB) = a (internal inverse angle). | AC | = | CD | Since | AB | = | AC | m (ACB) = 2a (outer angle). In the triangle ABC, m (ACB) = m (CBA) = 2a. Then 5a = 180 °, a = 36 °. x = 108 °.

Solution: | AB | = | AE |; m (AEB) = m (EBA) = b + 30 °, | AB | = | BD |; m (ADB) = m (BAD) = a + 24 ° A + 2b = 120 ° in triangle ABE, 2a + b = 132 ° in triangle USA. From the solution of a system of equations with two unknowns, b = 36 °. x = 42 ° is calculated.

Solution: Since a + b = 66 ° (outer angle), 2x + 2 (a + b) = 180 ° in triangle ABP; 2x = 180 ° -132 °, x = 24 °.

Solution: From top to bottom, in the triangle, equilateral angle, outer angle, equilateral angle in triangle, equilateral angle in triangle, outer angle in triangle, equilateral angle are written. The interior angles of the triangle are added together and the angle x is calculated