In an isosceles trapezoid ABCD (BC parallel to AD), BC = 9cm, AD = 21cm, AB = 10cm. Find: sin, cos, tg of angle D

Equal is a trapezoid in which the sides are equal, and the angles at the bases are equal:
AB = CD = 10 cm;
∠А = ∠D;
∠В = ∠С.
Let’s draw the height of the CH and consider the triangle ΔCDH. This triangle is right-angled with a right angle ∠Н.
In order to find the sine, cosine, tangent of the angle ∠D, we find the length of the CH.
CD ^ 2 = CH ^ 2 + HD ^ 2;
CH ^ 2 = CD ^ 2 – HD ^ 2;
HD = (AD – BC) / 2;
HD = (21 – 9) / 2 = 12/2 = 6 cm.
CH ^ 2 = 10 ^ 2 – 6 ^ 2 = 100 – 36 = 64;
CH = √64 = 8 cm.
The cosine of an acute angle of a right-angled triangle is the ratio of the adjacent leg to the hypotenuse:
cos D = CH / CD;
cos D = 8/10 = 0.8;
The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse:
sin D = HD / CD;
sin D = 6/10 = 0.6;
The tangent of an acute angle of a right triangle is the ratio of the opposite leg to the adjacent one:
tg D = CH / HD;
tg D = 8/6 = 1.333.
Answer: cos D = 0.8, sin D = 0.6, tg D = 1.333.